Unit 1: Limit and Continuity [7 Hours]

This unit introduces the limit as the value a function approaches, the epsilon-delta condition that makes the idea precise, and one-sided limits. It then develops the algebra of limits, the definition and conditions of continuity, and the continuity of algebraic, trigonometric, and exponential functions. Worked examples cover standard limits, piecewise continuity, trigonometric limits, and L'Hôpital's rule as tested in the supplied questions.

Tested Topics

1. Standard limits \((e^x-1)/x\), \((a^x-1)/x\), \((x^n-a^n)/(x-a)\) — 6x (2024, 2024, 2024, 2023, 2023, 2023)
2. Continuity / discontinuity of a piecewise-defined function and making it continuous — 4x (2024, 2024, 2023, 2023)
3. Evaluation of limits by substitution / algebraic simplification — 3x (2024, 2023, 2023)
4. Trigonometric limits \((\sin x / x\), \(\tan x\) variants, \(\sin x^\circ / x)\) — 3x (2024, 2023, 2023)
5. L'Hôpital's Rule — application to evaluate indeterminate limits — 3x (2024, 2023, 2023)
6. Indeterminate forms — definition — 1x (2023)
7. Limit of a function at infinity — 1x (2023)
8. Rolle's theorem condition (continuity on closed interval) — 1x (2023)

1.1 Definition of Limit Including Epsilon-Delta Condition, Right and Left Hand Limit, and Its Interpretation

Definition

Let \(f\) be defined near \(x=a\), but not necessarily at \(a\). We say that the limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\), written

\[ \lim_{x \to a} f(x) = L, \]

if \(f(x)\) can be made as close to \(L\) as we wish by taking \(x\) sufficiently close to \(a\), but not equal to \(a\).
The epsilon-delta condition makes “as close as we wish” precise: for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that

\[ 0 < |x-a| < \delta \quad \Rightarrow \quad |f(x)-L| < \varepsilon. \]

Detailed Explanation

Why limits are needed

A function may be undefined at \(a\), or may have a value at \(a\) different from the trend nearby. The limit describes the trend of \(f(x)\) near \(a\), not the value at \(a\).

Meaning of the epsilon-delta symbols

• \(\varepsilon\) (epsilon) is the allowed error between \(f(x)\) and \(L\).
• \(\delta\) (delta) is how close \(x\) must be to \(a\).
• The condition \(0 < |x-a|\) excludes \(x=a\), so the value \(f(a)\) does not affect the limit.

Right-hand and left-hand limits

The left-hand limit is \(\lim_{x \to a^-} f(x)\), using \(x < a\). The right-hand limit is \(\lim_{x \to a^+} f(x)\), using \(x > a\). The two-sided limit exists exactly when both one-sided limits exist and are equal:

\[ \lim_{x \to a} f(x)=L \quad \Longleftrightarrow \quad \lim_{x \to a^-} f(x)=\lim_{x \to a^+} f(x)=L. \]

Formula and Conditions

For a linear function \(f(x)=mx+c\),

\[ \lim_{x \to a} (mx+c)=ma+c. \]

To prove \(\lim_{x \to a} f(x)=L\) by epsilon-delta, choose \(\delta\) in terms of \(\varepsilon\), usually by simplifying \(|f(x)-L|\) into a multiple of \(|x-a|\).

Numerical Example 1: Epsilon-delta verification of a linear limit

Q. Prove that \(\lim_{x \to 2} (3x+1)=7\) using the epsilon-delta definition.

Solution

Step 1: Write the required inequality.

We need

\[ |(3x+1)-7| < \varepsilon. \]

Step 2: Simplify in terms of \(|x-2|\).

\[ |(3x+1)-7| = |3x-6| = 3|x-2|. \]

Step 3: Choose \(\delta\).

We want \(3|x-2| < \varepsilon\), so take

\[ \delta = \frac{\varepsilon}{3}. \]

Step 4: Verify the implication.

If \(0 < |x-2| < \delta = \varepsilon/3\), then

\[ |(3x+1)-7| = 3|x-2| < 3\cdot \frac{\varepsilon}{3} = \varepsilon. \]

∴ \(\lim_{x \to 2}(3x+1)=7\).

Check

Direct substitution gives \(3(2)+1=7\), so the limit value is correct. The epsilon-delta proof confirms the same value by the formal definition.

Explanation of Each Step

1. Step 1 — The definition requires bounding the distance between \(f(x)\) and \(L=7\).
2. Step 2 — Algebra simplifies the distance to \(3|x-2|\), which involves the distance from \(x\) to \(2\).
3. Step 3 — Dividing \(\varepsilon\) by 3 makes the final expression smaller than \(\varepsilon\).
4. Step 4 — The chosen \(\delta\) works for every positive \(\varepsilon\), so the limit is proved.

Numerical Example 2: Right-hand and left-hand limits of a piecewise function

Q. Let \(f(x)=x+1\) for \(x<2\), and \(f(x)=3x-5\) for \(x\ge 2\). Find \(\lim_{x \to 2^-}f(x)\), \(\lim_{x \to 2^+}f(x)\), and decide whether \(\lim_{x \to 2}f(x)\) exists.

Solution

Step 1: Left-hand limit.

For \(x<2\), \(f(x)=x+1\). Thus

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-}(x+1)=2+1=3. \]

Step 2: Right-hand limit.

For \(x>2\), \(f(x)=3x-5\). Thus

\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+}(3x-5)=3(2)-5=1. \]

Step 3: Compare.

Since \(3\neq 1\), the two one-sided limits are not equal.

∴ \(\lim_{x \to 2^-}f(x)=3\), \(\lim_{x \to 2^+}f(x)=1\), and \(\lim_{x \to 2}f(x)\) does not exist.

Check

A graph jumps from height 3 on the left to height 1 on the right at \(x=2\), so no single two-sided limit can exist.

Explanation of Each Step

1. Step 1 — Use the left branch because \(x\) approaches 2 from below.
2. Step 2 — Use the right branch because \(x\) approaches 2 from above.
3. Step 3 — A two-sided limit exists only if both one-sided limits agree. They do not, so the two-sided limit fails.

1.2 Algebraic Properties of Limit

Definition

The algebra of limits is the set of rules that allow limits to be split, combined, and simplified when the separate limits exist. If \(\lim_{x\to a}f(x)=L\) and \(\lim_{x\to a}g(x)=M\), then sums, products, quotients, powers, and roots of limits follow the expected rules.

Detailed Explanation

These rules let us evaluate a complicated limit by breaking it into simpler known limits. They apply only when the individual limits exist and, for quotients, when the denominator limit is not zero.

Basic properties

• Sum: \(\lim_{x\to a}[f(x)+g(x)]=L+M\).
• Difference: \(\lim_{x\to a}[f(x)-g(x)]=L-M\).
• Product: \(\lim_{x\to a}[f(x)g(x)]=LM\).
• Constant multiple: \(\lim_{x\to a}[cf(x)]=cL\).
• Quotient: \(\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{L}{M}\), provided \(M\neq 0\).
• Power: \(\lim_{x\to a}[f(x)]^n=L^n\).
• Root: \(\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{L}\), where the root is defined.

Formula and Conditions

Standard limits used repeatedly:

\[ \lim_{x\to 0}\frac{e^x-1}{x}=1,\qquad \lim_{x\to 0}\frac{a^x-1}{x}=\ln a\ (a>0), \]

\[ \lim_{x\to a}\frac{x^n-a^n}{x-a}=n a^{n-1}\quad \text{(for rational }n\text{ where the expression is defined)}. \]

If substitution gives \(0/0\), factor, rationalize, or use a standard limit. The quotient rule cannot be used when the denominator limit is zero.

Procedure

1. Try direct substitution.
2. If the result is defined, that is the limit.
3. If \(0/0\) appears, simplify algebraically: factor, cancel, rationalize, or use a standard limit.
4. Apply the algebraic properties to combine the simplified pieces.

Numerical Example 1: Evaluation by algebraic simplification

Q. Evaluate \(\displaystyle \lim_{x\to 2}\frac{x^2-4}{x-2}\).

Solution

Step 1: Substitute.

\[ \frac{2^2-4}{2-2}=\frac{0}{0}, \]

an indeterminate form, so simplification is needed.

Step 2: Factor and cancel.

\[ \frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2,\qquad x\neq 2. \]

Step 3: Take the limit.

\[ \lim_{x\to 2}(x+2)=2+2=4. \]

∴ \(\displaystyle \lim_{x\to 2}\frac{x^2-4}{x-2}=4\).

Check

Values near \(x=2\), such as \(x=2.1\), give \(\frac{4.41-4}{0.1}=4.1\), close to 4. The cancellation is valid because \(x\neq 2\) in the limit process.

Explanation of Each Step

1. Step 1 — Direct substitution gives \(0/0\), which is not a value; it signals a common factor.
2. Step 2 — Factor the numerator as a difference of squares and cancel the common factor \(x-2\), allowed because \(x\neq 2\).
3. Step 3 — The simplified expression is continuous at \(x=2\), so substitute.

Numerical Example 2: Standard exponential and power limits

Q. Evaluate (a) \(\displaystyle \lim_{x\to 0}\frac{e^{3x}-1}{x}\), (b) \(\displaystyle \lim_{x\to a}\frac{x^5-a^5}{x-a}\).

Solution

Step 1: Part (a). Write \(e^{3x}-1=(e^{3x}-1)\). Let \(u=3x\). As \(x\to 0\), \(u\to 0\). Then

\[ \frac{e^{3x}-1}{x}=3\cdot \frac{e^{3x}-1}{3x}=3\cdot \frac{e^u-1}{u}. \]

Using \(\lim_{u\to 0}\frac{e^u-1}{u}=1\),

\[ \lim_{x\to 0}\frac{e^{3x}-1}{x}=3(1)=3. \]

Step 2: Part (b). Use the standard power limit with \(n=5\):

\[ \lim_{x\to a}\frac{x^5-a^5}{x-a}=5a^{5-1}=5a^4. \]

∴ (a) \(3\), (b) \(5a^4\).

Check

For (a), expand \(e^{3x}=1+3x+\frac{(3x)^2}{2}+\cdots\), so \(\frac{e^{3x}-1}{x}=3+\frac{9x}{2}+\cdots\to 3\). For (b), factor \(x^5-a^5=(x-a)(x^4+x^3a+x^2a^2+xa^3+a^4)\), giving \(5a^4\) at \(x=a\).

Explanation of Each Step

1. Step 1 — Multiply and divide by 3 to create the standard form \((e^u-1)/u\).
2. Step 2 — The standard limit \((x^n-a^n)/(x-a)\to n a^{n-1}\) is the derivative of \(x^n\) at \(a\).

1.3 Definition and Conditions of Continuity and Discontinuity

Definition

A function \(f\) is continuous at \(x=a\) if three conditions hold:

1. \(f(a)\) is defined,
2. \(\lim_{x\to a}f(x)\) exists,
3. \(\lim_{x\to a}f(x)=f(a)\).

If any condition fails, \(f\) is discontinuous at \(a\).

Detailed Explanation

Continuity means the graph has no hole, jump, or vertical break at \(a\). The limit from both sides must agree with the actual function value.

Types of discontinuity

• Removable discontinuity: the limit exists but \(f(a)\) is missing or different. Redefine \(f(a)\) to make it continuous.
• Jump discontinuity: left and right limits exist but are different.
• Infinite discontinuity: the function tends to \(\pm\infty\) near \(a\).
• Oscillatory discontinuity: the function does not settle to any limit.

Formula and Conditions

For a piecewise function, check continuity at a boundary point \(a\) by computing

\[ \lim_{x\to a^-}f(x),\qquad \lim_{x\to a^+}f(x),\qquad f(a). \]

The function is continuous at \(a\) exactly when all three are equal.

Procedure

1. Find the domain and identify boundary points.
2. Compute the left-hand limit.
3. Compute the right-hand limit.
4. Compute \(f(a)\).
5. Compare the three values. If they agree, continuous; otherwise discontinuous.
6. For a removable discontinuity, set \(f(a)\) equal to the common limit.

Numerical Example 1: Piecewise continuity and making it continuous

Q. A function \(f(x)\) is defined as \(f(x)=2x+3\) for \(x<1\), \(f(x)=4\) for \(x=1\), \(f(x)=6x-1\) for \(x>1\). Is the function continuous at \(x=1\)? If not, how can you make it continuous at \(x=1\)?

Solution

Step 1: Left-hand limit.

\[ \lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}(2x+3)=2(1)+3=5. \]

Step 2: Right-hand limit.

\[ \lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}(6x-1)=6(1)-1=5. \]

Step 3: Function value.

\[ f(1)=4. \]

Step 4: Compare.

The one-sided limits are both 5, so \(\lim_{x\to 1}f(x)=5\). But \(f(1)=4\neq 5\). Therefore \(f\) is not continuous at \(x=1\).

Step 5: Make it continuous.

Redefine \(f(1)=5\). Then \(\lim_{x\to 1}f(x)=f(1)=5\).

∴ The function is discontinuous at \(x=1\); redefine \(f(1)=5\) to make it continuous.

Check

After redefinition, the graph meets at height 5 from both sides and passes through \((1,5)\).

Explanation of Each Step

1. Step 1 — Use the left branch \(2x+3\) because \(x<1\).
2. Step 2 — Use the right branch \(6x-1\) because \(x>1\).
3. Step 3 — The value at \(x=1\) is separately given as 4.
4. Step 4 — Continuity requires the limit and function value to be equal. They are 5 and 4, so it fails.
5. Step 5 — Since the limit exists, the discontinuity is removable by changing \(f(1)\).

Numerical Example 2: Removable discontinuity

Q. Find the value of \(k\) that makes \(f(x)=\frac{x^2-1}{x-1}\) for \(x\neq 1\), and \(f(1)=k\), continuous at \(x=1\).

Solution

Step 1: Simplify for \(x\neq 1\).

\[ \frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1. \]

Step 2: Find the limit.

\[ \lim_{x\to 1}\frac{x^2-1}{x-1}=\lim_{x\to 1}(x+1)=2. \]

Step 3: Set the function value equal to the limit.

\[ k=2. \]

∴ \(k=2\).

Check

With \(k=2\), \(f(1)=2\) and the limit as \(x\to 1\) is 2, so the function is continuous at \(x=1\).

Explanation of Each Step

1. Step 1 — Factor the numerator and cancel \(x-1\), valid because \(x\neq 1\).
2. Step 2 — Substitute \(x=1\) into the simplified expression \(x+1\).
3. Step 3 — Continuity requires \(f(1)\) to equal the limit.

1.4 Continuity of Algebraic, Trigonometric, and Exponential Functions, Examples, and Counterexamples

Definition

Polynomials, rational functions where the denominator is nonzero, trigonometric functions, and exponential functions are continuous on their domains. Their limits can often be found by substitution. Special limits are needed when substitution gives \(0/0\).

Detailed Explanation

Algebraic functions are built from polynomials using addition, subtraction, multiplication, division, and roots. They are continuous wherever they are defined. Trigonometric functions \(\sin x\), \(\cos x\), \(\tan x\), etc. are continuous on their domains. Exponential functions \(e^x\), \(a^x\) are continuous for all real \(x\).

Important trigonometric limits

When \(x\) is in radians,

\[ \lim_{x\to 0}\frac{\sin x}{x}=1,\qquad \lim_{x\to 0}\frac{\tan x}{x}=1. \]

When \(x^\circ\) means \(x\) degrees, convert: \(x^\circ=\frac{\pi x}{180}\), so

\[ \lim_{x\to 0}\frac{\sin x^\circ}{x}=\frac{\pi}{180}. \]

Formula and Conditions

For exponential limits,

\[ \lim_{x\to 0}\frac{e^x-1}{x}=1,\qquad \lim_{x\to 0}\frac{a^x-1}{x}=\ln a. \]

For trigonometric limits, the radian measure is required unless degrees are explicitly converted. A counterexample to continuity is any function with a jump, hole, or infinite break on its domain.

Numerical Example 1: Trigonometric limits

Q. Evaluate (a) \(\displaystyle \lim_{x\to 0}\frac{\sin x}{x}\), (b) \(\displaystyle \lim_{x\to 0}\frac{\sin x^\circ}{x}\), (c) \(\displaystyle \lim_{x\to 0}\frac{\tan x-\sin x}{x^3}\).

Solution

Step 1: Part (a).

This is the standard trigonometric limit:

\[ \lim_{x\to 0}\frac{\sin x}{x}=1. \]

Step 2: Part (b).

Convert degrees to radians: \(x^\circ=\frac{\pi x}{180}\). Then

\[ \frac{\sin x^\circ}{x}=\frac{\sin\left(\frac{\pi x}{180}\right)}{x} =\frac{\pi}{180}\cdot \frac{\sin\left(\frac{\pi x}{180}\right)}{\frac{\pi x}{180}}. \]

As \(x\to 0\), the inner angle tends to 0, so the fraction tends to 1. Hence

\[ \lim_{x\to 0}\frac{\sin x^\circ}{x}=\frac{\pi}{180}. \]

Step 3: Part (c).

\[ \tan x-\sin x=\frac{\sin x}{\cos x}-\sin x =\sin x\left(\frac{1}{\cos x}-1\right) =\sin x\frac{1-\cos x}{\cos x}. \]

Thus

\[ \frac{\tan x-\sin x}{x^3} =\frac{\sin x}{x}\cdot \frac{1-\cos x}{x^2}\cdot \frac{1}{\cos x}. \]

Using \(\frac{\sin x}{x}\to 1\), \(\frac{1-\cos x}{x^2}\to \frac12\), and \(\cos x\to 1\),

\[ \lim_{x\to 0}\frac{\tan x-\sin x}{x^3}=1\cdot \frac12\cdot 1=\frac12. \]

∴ (a) \(1\), (b) \(\frac{\pi}{180}\), (c) \(\frac12\).

Check

For (c), using Taylor series: \(\tan x=x+\frac{x^3}{3}+\cdots\), \(\sin x=x-\frac{x^3}{6}+\cdots\), so difference \(\approx \frac{x^3}{2}\), giving \(\frac12\).

Explanation of Each Step

1. Step 1 — Standard limit in radians.
2. Step 2 — Convert degrees to radians and scale to the standard sine limit.
3. Step 3 — Factor \(\sin x\), use the standard limits for \(\sin x/x\) and \((1-\cos x)/x^2\).

Numerical Example 2: Exponential continuity and limits

Q. Evaluate \(\displaystyle \lim_{x\to 0}\frac{e^{2x}-1}{x}\) and \(\displaystyle \lim_{x\to 0}\frac{a^x-1}{x}\).

Solution

Step 1: First limit.

\[ \frac{e^{2x}-1}{x}=2\cdot \frac{e^{2x}-1}{2x}. \]

Let \(u=2x\). As \(x\to 0\), \(u\to 0\). Hence

\[ \lim_{x\to 0}\frac{e^{2x}-1}{x}=2\lim_{u\to 0}\frac{e^u-1}{u}=2(1)=2. \]

Step 2: Second limit.

Using the standard exponential limit,

\[ \lim_{x\to 0}\frac{a^x-1}{x}=\ln a\qquad (a>0). \]

∴ The limits are \(2\) and \(\ln a\).

Check

For \(a=e\), \(\ln e=1\), matching the first standard limit. For \(a=10\), the limit is \(\ln 10\approx 2.3026\).

Explanation of Each Step

1. Step 1 — Multiply and divide by 2 to create \((e^u-1)/u\).
2. Step 2 — Use the known limit \((a^x-1)/x\to \ln a\), which follows from \(a^x=e^{x\ln a}\).

Extra: Topics Asked in Exams But Not in Syllabus

The supplied past questions include L'Hôpital's rule and indeterminate forms, which are not listed in the supplied syllabus. They are covered here so the exam questions can be solved.

L'Hôpital's Rule and Indeterminate Forms

Definition

Indeterminate forms are expressions such as

\[ \frac00,\quad \frac{\infty}{\infty},\quad 0\cdot\infty,\quad \infty-\infty,\quad 1^\infty,\quad 0^0,\quad \infty^0. \]

They are called indeterminate because their limiting value depends on the functions involved, not on the form alone.

L'Hôpital's rule states that if \(\lim_{x\to a}\frac{f(x)}{g(x)}\) gives \(0/0\) or \(\infty/\infty\), and \(f,g\) are differentiable near \(a\), then

\[ \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}, \]

provided the latter limit exists or is \(\pm\infty\).

Formula and Conditions

Use L'Hôpital's rule only after checking that the form is \(0/0\) or \(\infty/\infty\). Do not apply it to a determinate form. Differentiate numerator and denominator separately; do not use the quotient rule.

Numerical Example 1: L'Hôpital's rule for a simple indeterminate limit

Q. Using L'Hôpital's rule, evaluate \(\displaystyle \lim_{x\to 0}\frac{x^2}{4\sin x}\).

Solution

Step 1: Check the form.

\[ \frac{0^2}{4\sin 0}=\frac00. \]

Step 2: Apply L'Hôpital's rule.

\[ \lim_{x\to 0}\frac{x^2}{4\sin x} =\lim_{x\to 0}\frac{2x}{4\cos x} =\lim_{x\to 0}\frac{x}{2\cos x}. \]

Step 3: Substitute.

\[ \frac{0}{2\cos 0}=\frac{0}{2}=0. \]

∴ \(\displaystyle \lim_{x\to 0}\frac{x^2}{4\sin x}=0\).

Check

For small \(x\), \(\sin x\approx x\), so \(\frac{x^2}{4\sin x}\approx \frac{x}{4}\to 0\).

Explanation of Each Step

1. Step 1 — Substitution gives \(0/0\), so L'Hôpital's rule is allowed.
2. Step 2 — Differentiate numerator \(x^2\) to \(2x\), denominator \(4\sin x\) to \(4\cos x\).
3. Step 3 — The new expression is defined at \(x=0\), so substitute directly.

Numerical Example 2: L'Hôpital's rule applied twice

Q. Evaluate \(\displaystyle \lim_{x\to 0}\frac{xe^x-\ln(1+x)}{x^2}\).

Solution

Step 1: Check the form.

At \(x=0\): numerator \(0\cdot e^0-\ln 1=0\), denominator \(0\), so \(0/0\).

Step 2: First application.

\[ \lim_{x\to 0}\frac{xe^x-\ln(1+x)}{x^2} =\lim_{x\to 0}\frac{e^x+xe^x-\frac{1}{1+x}}{2x}. \]

At \(x=0\), numerator \(=1+0-1=0\), denominator \(=0\), still \(0/0\).

Step 3: Second application.

Differentiate numerator: \(\frac{d}{dx}(e^x+xe^x-\frac{1}{1+x}) = e^x(2+x)+\frac{1}{(1+x)^2}\). Differentiate denominator: \(2\). Thus

\[ \lim_{x\to 0}\frac{e^x(2+x)+\frac{1}{(1+x)^2}}{2} =\frac{1(2+0)+1}{2} =\frac{3}{2}. \]

∴ \(\displaystyle \lim_{x\to 0}\frac{xe^x-\ln(1+x)}{x^2}=\frac32\).

Check

Using series: \(xe^x=x+x^2+\frac{x^3}{2}+\cdots\), \(\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\). Difference \(=\frac{3x^2}{2}+\cdots\), so limit \(=\frac32\).

Explanation of Each Step

1. Step 1 — Confirm the indeterminate form \(0/0\).
2. Step 2 — Apply L'Hôpital's rule; the result is still \(0/0\).
3. Step 3 — Apply L'Hôpital's rule again; now the limit is determinate and substitution gives \(3/2\).

Numerical Example 3: L'Hôpital's rule with a parameter

Q. Evaluate \(\displaystyle \lim_{x\to \theta}\frac{x\cos\theta-\theta\cos x}{x-\theta}\).

Solution

Step 1: Check the form.

At \(x=\theta\): numerator \(\theta\cos\theta-\theta\cos\theta=0\), denominator \(0\), so \(0/0\).

Step 2: Apply L'Hôpital's rule.

\[ \lim_{x\to \theta}\frac{x\cos\theta-\theta\cos x}{x-\theta} =\lim_{x\to \theta}\frac{\cos\theta+\theta\sin x}{1}. \]

Step 3: Substitute.

\[ \cos\theta+\theta\sin\theta. \]

∴ \(\displaystyle \lim_{x\to \theta}\frac{x\cos\theta-\theta\cos x}{x-\theta}=\cos\theta+\theta\sin\theta\).

Check

If \(\theta=0\), the limit becomes \(\lim_{x\to 0}\frac{x}{x}=1\), and the formula gives \(\cos 0+0=1\).

Explanation of Each Step

1. Step 1 — Substitution gives \(0/0\).
2. Step 2 — Differentiate numerator: derivative of \(x\cos\theta\) is \(\cos\theta\), derivative of \(-\theta\cos x\) is \(\theta\sin x\). Denominator derivative is 1.
3. Step 3 — Substitute \(x=\theta\).

Exam Questions

1. For all rational values of \(n\), \(\lim_{x \to a}\left(\frac{x^n-a^n}{x-a}\right)\) is equal to: (2023)
2. The value of \(\lim_{x \to 0}\left(\frac{e^x-1}{x}\right)\) is equal to: (2023)
3. The limit of the function \(f(x) = \frac{e^x - 1}{x}\) as \(x \to 0\) — A.−1, B.1, C.0, D.does not exist (2024)
4. \(\lim_{x \to 0} \frac{e^{2x} - 1}{x} =\) (2023)
5. The value of \(\lim_{x \to 0} \frac{a^x - 1}{x}\) is equal to (2023)
6. The function \(y = \frac{x^2 - 4}{x - 2}\) is discontinuous at — options (2023)
7. Using L'Hôpital's rule, the value of \(\lim_{x \to 0} \frac{x^2}{4\sin x}\) (2023)
8. Limit of the function \(\lim_{x \to 0} \frac{\sin x^°}{x}\) (2023)
9. Define Indeterminate forms. (2023)
10. Evaluate: \(\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}\) (2023)
11. A function \(f(x)\) is defined as \(f(x)=2x+3\) for \(x<1\), \(f(x)=4\) for \(x=1\), \(f(x)=6x-1\) for \(x>1\). Is the function continuous at \(x=1\)? If not, how can you make it continuous at \(x=1\)? (2023 / 2024)
12. Evaluate the limit \(\lim_{x \to \theta}\frac{x\cos\theta - \theta\cos x}{x - \theta}\). (2023 / 2023)
13. (a) State L'Hospital Rule. Use it to evaluate \(\lim_{x \to 0}\left(\frac{xe^x-\ln(1+x)}{x^2}\right)\) (2023)
14. If a function \(f(x) = \frac{x^2-1}{x-1}\) for \(x\neq1\), \(k\) for \(x=1\) is continuous at \(x=1\), then value of \(k\) is (2023)

Whole Chapter Summary

1.1 Limit: The limit describes the trend of \(f(x)\) near \(a\). Epsilon-delta gives the precise definition, and one-sided limits decide whether a two-sided limit exists.

1.2 Algebra of limits: Limits combine over sums, products, quotients, powers, and roots. Standard limits handle \(0/0\) forms involving exponentials and powers.

1.3 Continuity: A function is continuous at \(a\) when \(f(a)\), the limit, and the one-sided limits all agree. Discontinuities are removable, jump, infinite, or oscillatory.

1.4 Continuity of functions: Algebraic, trigonometric, and exponential functions are continuous on their domains. Special trigonometric and exponential limits are essential.

Extra: Indeterminate forms and L'Hôpital's rule extend limit evaluation to \(0/0\) and \(\infty/\infty\) cases.

Key Formulas and Points Sheet

Last-Minute Revision Sheet