If lim x→0 (ax - (e^(4x) - 1)) / (ax(e^(4x) - 1)) exists and is equal to b, then the value of a - 2b

Q: Wait, why can't I just use the Taylor series expansion for $e^{4x}$ right from the start?

You absolutely can! In fact, using the Taylor expansion $e^{4x} = 1 + 4x + \frac{(4x)^2}{2!} + \dots$ is often more elegant than L'Hopital's Rule. It turns the limit into a polynomial ratio, which makes the behavior near $x=0$ crystal clear. Both methods are valid, but L'Hopital's is a great way to practice your differentiation skills.

Q: Why did we set the numerator to zero in step 5? Is that a rule?

It is a consequence of the existence of a finite limit. If the denominator approaches zero and the numerator approaches a non-zero constant, the limit would diverge to infinity. Since the problem states the limit exists and equals $b$, the numerator must also approach zero to create the necessary indeterminate form that allows for a finite value.

Q: Can I use the standard limit $\lim_{x \to 0} \frac{e^{kx}-1}{x} = k$ directly?

Yes, that is a powerful tool! Recognizing that $e^{4x}-1 \approx 4x$ is essentially using that standard limit. It simplifies the denominator significantly and helps you avoid messy product rule differentiation.

Q: Why did we apply L'Hopital's Rule twice?

After the first application, we were left with a variable $x$ in the denominator. If we had tried to evaluate the limit then, we would still have a $0/0$ form or an undefined expression. We apply it again to 'peel away' the layers of the indeterminate form until the denominator becomes a non-zero constant.

Q: Is it possible for $a$ to be something other than 4?

If $a$ were any value other than 4, the numerator would not approach zero at the same rate as the denominator, causing the limit to blow up to infinity. The constraint that the limit exists is exactly what 'locks' the value of $a$ to 4.

Question

If $\lim_{x	o 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)}$ exists and is equal to $b$, then the value of $a-2b$ is

Sigma Percentile · Accepted Answer

### The Dance of the Indeterminate Form

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that, at first glance, looks like a standard calculus exercise.

But beneath the surface of this limit, there lies a beautiful interplay between algebraic structure and the fundamental nature of functions as they approach the origin. We are tasked with finding the value of $a - 2b$ given that the limit

$$\lim_{x	o 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)} = b$$

exists.

### Phase 1: The Diagnostic Check

Whenever you see a limit, your first instinct should be to test the waters. What happens when $x$ actually reaches $0$?

Substituting $x=0$ into our expression, we find the numerator becomes $a(0) - (e^0 - 1) = 0 - 0 = 0$, and the denominator becomes $a(0)(e^0 - 1) = 0 \cdot 0 = 0$.

We have arrived at the classic $\frac{0}{0}$ indeterminate form. This is not a dead end; it is an invitation to strip away the layers of complexity.

### Phase 2: Simplifying the Landscape

Looking at the denominator $ax(e^{4x}-1)$, we could dive straight into the product rule, but that is a path to algebraic exhaustion. Instead, let us use our intuition about standard limits.

We know that near the origin, the exponential function $e^{4x}$ behaves linearly. Specifically, the standard limit $\lim_{x	o 0} \frac{e^{kx}-1}{kx} = 1$ tells us that $e^{4x}-1 \approx 4x$.

By replacing the exponential term with its linear approximation, our denominator simplifies to $ax(4x) = 4ax^2$. Now, our limit looks much more manageable:

$$\lim_{x	o 0} \frac{ax-(e^{4x}-1)}{4ax^2} = b$$

### Phase 3: The Power of L'Hopital's Rule

Even with the simplification, we are still staring at a $\frac{0}{0}$ form. It is time to call upon the heavy artillery: L'Hopital's Rule.

By differentiating the numerator and the denominator with respect to $x$, we get:

$$\lim_{x	o 0} \frac{a - 4e^{4x}}{8ax} = b$$

Here is where the magic happens. Look closely at the denominator $8ax$. As $x$ approaches $0$, the denominator vanishes.

For the entire limit to result in a finite value $b$, the numerator must also vanish at $x=0$. If it didn't, we would have a non-zero constant divided by zero, which would explode to infinity.

Therefore, we must have $a - 4e^{4(0)} = 0$, which immediately gives us **$a = 4$**.

### Phase 4: The Final Resolution

With $a=4$ in hand, the limit becomes:

$$\lim_{x	o 0} \frac{4 - 4e^{4x}}{32x} = b$$

We are still at $\frac{0}{0}$, so we apply L'Hopital's Rule one final time. Differentiating the numerator gives us $-16e^{4x}$, and the denominator becomes $32$.

Now, the limit is simply:

$$\lim_{x	o 0} \frac{-16e^{4x}}{32} = b$$

Substituting $x=0$, we find **$b = -\frac{1}{2}$**.

We have successfully navigated the trap! With $a=4$ and $b=-\frac{1}{2}$, the final calculation is straightforward:

$$a - 2b = 4 - 2\left(-\frac{1}{2}ight) = 4 + 1 = \mathbf{5}$$

Remember, in the world of JEE, limits are not just about plugging in numbers; they are about understanding how functions compete to reach zero. You have mastered the logic, and that is the true victory.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 0} \frac{a x - ( e ^{4 x} - 1 )}{a x ( e ^{4 x} - 1 )}$ exists and is equal to $b$ , then the value of $a - 2 b$ is

Enter Numerical Value:

Visualized Solution

Analyzing the Limit Form

Simplifying the Denominator

Standard Limit Approximation

Rewriting the Limit

Applying L'Hopital's Rule

Condition for Finite Limit

Solving for Constant $a$

Substituting $a$ and Re-evaluating

Second Application of L'Hopital's Rule

Finding the Value of $b$

Final Calculation

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

The Dance of the Indeterminate Form

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→0​ax(e4x−1)ax−(e4x−1)​ exists and is equal to b, then the value of a−2b is