Why can't we just plug in $x=0$ directly?

Because the function is undefined at $x=0$ due to division by zero. We are looking for the limit, which tells us where the function *should* be to make it continuous. It is like finding the missing piece of a puzzle.

Why do we have to use L'Hopital's Rule twice?

Because the first derivative still yields an indeterminate form of $\frac{0}{0}$. It is a sign that the function's behavior near zero is more complex than a simple linear slope, requiring a deeper look at the rate of change.

Is there a faster way to solve this?

Yes, using the Taylor Series expansion for $e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \dots$. This is a powerful alternative for advanced students that bypasses the need for repeated differentiation.

JEE Main 2007

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The function $f : R / {0} \to R$ given by $f (x) = \frac{1}{x} - \frac{2}{e ^{2 x} - 1}$ can be made continuous at $x = 0$ by defining $f (0)$ as

Select Answer:

Visualized Solution

Understanding the Discontinuity

Given function: $f (x) = \frac{1}{x} - \frac{2}{e ^{2 x} - 1}$
At $x = 0$ , the function is undefined due to division by zero.
Graphically, this creates a hole at $x = 0$ .

Condition for Continuity

For $f (x)$ to be continuous at $x = 0$ :
$f (0) = lim_{x \to 0} f (x)$
We must evaluate: $lim_{x \to 0} (\frac{1}{x} - \frac{2}{e ^{2 x} - 1})$

Algebraic Simplification

Take the Least Common Multiple (LCM) of the denominators.
$f (0) = lim_{x \to 0} \frac{e ^{2 x} - 1 - 2 x}{x ( e ^{2 x} - 1 )}$

Checking the Limit Form

Substitute $x = 0$ into the combined expression.
Numerator: $e^{0} - 1 - 2 (0) = 1 - 1 - 0 = 0$
Denominator: $0 (e^{0} - 1) = 0 (0) = 0$
The limit is in the $\frac{0}{0}$ indeterminate form.

L'Hopital's Rule: First Derivative

Apply L'Hopital's Rule: Differentiate numerator and denominator.
$\frac{d}{d x} (Numerator) = 2 e^{2 x} - 2$
$\frac{d}{d x} (Denominator) = (e^{2 x} - 1) + x (2 e^{2 x})$
New limit: $lim_{x \to 0} \frac{2 e ^{2 x} - 2}{e ^{2 x} - 1 + 2 x e ^{2 x}}$

Re-evaluating the Limit

Substitute $x = 0$ into the new expression.
Numerator: $2 e^{0} - 2 = 2 (1) - 2 = 0$
Denominator: $(e^{0} - 1) + 0 = (1 - 1) + 0 = 0$
The limit is still in the $\frac{0}{0}$ indeterminate form.

L'Hopital's Rule: Second Derivative

Differentiate numerator and denominator again.
$\frac{d}{d x} (Numerator) = 4 e^{2 x}$
$\frac{d}{d x} (Denominator) = 2 e^{2 x} + (2 e^{2 x} + 4 x e^{2 x}) = 4 e^{2 x} + 4 x e^{2 x}$
New limit: $lim_{x \to 0} \frac{4 e ^{2 x}}{4 e ^{2 x} + 4 x e ^{2 x}}$

Final Substitution

Substitute $x = 0$ into the final simplified limit.
$f (0) = \frac{4 e ^{0}}{4 e ^{0} + 4 ( 0 ) e ^{0}}$
$f (0) = \frac{4 ( 1 )}{4 ( 1 ) + 0}$
$f (0) = \frac{4}{4} = 1$

Filling the Hole

By defining $f (0) = 1$ , the function becomes continuous at $x = 0$ .
The hole in the graph at $(0, 1)$ is now filled.
Final Answer: $1$

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The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Solution Diagram

Analyzing the Problem

We are given the function:

f (x) = \frac{1}{x} - \frac{2}{e ^{2 x} - 1}

x = 0

, the function encounters an indeterminate form, as both terms approach infinity. To make the function continuous at

x = 0

, we must determine the value of the limit:

x \to 0 lim f (x) = x \to 0 lim (\frac{1}{x} - \frac{2}{e ^{2 x} - 1})

Simplifying the Expression

To evaluate this limit, we combine the terms into a single fraction by finding a common denominator:

f (x) = \frac{( e ^{2 x} - 1 ) - 2 x}{x ( e ^{2 x} - 1 )}

Substituting

x = 0

into this expression yields the indeterminate form

\frac{0}{0}

. This indicates that we must apply L'Hopital's Rule.

Applying L'Hopital's Rule

We differentiate the numerator and the denominator with respect to

x

Numerator derivative:

\frac{d}{d x} (e^{2 x} - 1 - 2 x) = 2 e^{2 x} - 2

Denominator derivative (using the product rule):

\frac{d}{d x} (x (e^{2 x} - 1)) = (e^{2 x} - 1) + x (2 e^{2 x})

The limit becomes:

x \to 0 lim \frac{2 e ^{2 x} - 2}{e ^{2 x} - 1 + 2 x e ^{2 x}}

Testing

x = 0

again results in

\frac{0}{0}

. We apply L'Hopital's Rule a second time.

Final Calculation

Differentiating the numerator and denominator once more:

Numerator:

\frac{d}{d x} (2 e^{2 x} - 2) = 4 e^{2 x}

Denominator:

\frac{d}{d x} (e^{2 x} - 1 + 2 x e^{2 x}) = 2 e^{2 x} + (2 e^{2 x} + 4 x e^{2 x}) = 4 e^{2 x} + 4 x e^{2 x}

Now, we evaluate the limit as

x

approaches

0

x \to 0 lim \frac{4 e ^{2 x}}{4 e ^{2 x} + 4 x e ^{2 x}} = \frac{4 ( 1 )}{4 ( 1 ) + 4 ( 0 ) ( 1 )} = \frac{4}{4} = 1

By defining $f (0) = 1$ , we successfully remove the discontinuity and make the function continuous at

x = 0

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The function $f : R / {0} \to R$ given by $f (x) = \frac{1}{x} - \frac{2}{e ^{2 x} - 1}$ can be made continuous at $x = 0$ by defining $f (0)$ as

Select Answer:

Visualized Solution

Understanding the Discontinuity

Condition for Continuity

Algebraic Simplification

Checking the Limit Form

L'Hopital's Rule: First Derivative

Re-evaluating the Limit

L'Hopital's Rule: Second Derivative

Final Substitution

Filling the Hole

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Problem

Simplifying the Expression

Applying L'Hopital's Rule

Final Calculation

Similar Questions

If the function $f (x) = \frac{t a n ( t a n x ) - s i n ( s i n x )}{t a n x - s i n x}$ is continuous at $x = 0$ , then $f (0)$ is equal to ________

Let $f : R \to R$ be such that $f (1) = 3$ and $f^{'} (1) = 6$ . Then $lim_{x \to 0} (\frac{f ( 1 + x )}{f ( 1 )})^{1/ x}$ equals

$lim_{x \to 0} \frac{e - ( 1 + 2 x ) ^{\frac{1}{2 x}}}{x}$ is equal to

Let $f : R \to R$ be a differentiable function satisfying $f^{'} (3) + f^{'} (2) = 0$ . Then $lim_{x \to 0} (\frac{1 + f ( 3 + x ) - f ( 3 )}{1 + f ( 2 - x ) - f ( 2 )})^{1/ x}$ is equal to

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Let $f : (- \infty, \infty) - {0} \to R$ be a differentiable function such that $f^{'} (1) = lim_{a \to \infty} a^{2} f (\frac{1}{a})$ . Then $lim_{a \to \infty} \frac{a ( a + 1 )}{2} tan^{- 1} (\frac{1}{a}) + a^{2} - 2 lo g_{e} a$ is equal to

Select Answer:

Visualized Solution

Understanding the Discontinuity

Condition for Continuity

Algebraic Simplification

Checking the Limit Form

L'Hopital's Rule: First Derivative

Re-evaluating the Limit

L'Hopital's Rule: Second Derivative

Final Substitution

Filling the Hole

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Problem

Simplifying the Expression

Applying L'Hopital's Rule

Final Calculation

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