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Limits Continuity and Differentiability question

2021 · 25 Feb · Shift 2 · Q47

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Limits Continuity and Differentiability
2021 · 25 Feb · Shift 2 · Q47

Limits Continuity and Differentiability question

2021 · 25 Feb · Shift 2 · Q47

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1

If

\mathop {\lim }\limits_{x \to 0} {{ax - ({e^{4x}} - 1)} \over {ax({e^{4x}} - 1)}}

exists and is equal to b, then the value of a

-

2b is

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 5

We need to evaluate

\lim_{x\to 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)}

and it is given that this limit exists and equals $b$ .

For the limit to be finite, the numerator must vanish to sufficiently high order as $x\to 0$ .

Use the expansion:

e^{4x}=1+4x+8x^2+\frac{32}{3}x^3+\cdots

So,

e^{4x}-1=4x+8x^2+\frac{32}{3}x^3+\cdots

Substitute into the numerator:

ax-(e^{4x}-1)=ax-\left(4x+8x^2+\frac{32}{3}x^3+\cdots\right)

=(a-4)x-8x^2-\frac{32}{3}x^3+\cdots

Now look at the denominator:

ax(e^{4x}-1)=ax\left(4x+8x^2+\frac{32}{3}x^3+\cdots\right)

=4ax^2+8ax^3+\cdots

Since the denominator is of order $x^2$ , for the limit to exist finitely, the numerator must also be at least of order $x^2$ .

Thus the coefficient of $x$ in the numerator must be zero:

a-4=0 \implies a=4

Now substitute $a=4$ into the expression:

\lim_{x\to 0} \frac{4x-(e^{4x}-1)}{4x(e^{4x}-1)}

Using the expansion,

4x-(e^{4x}-1)=4x-\left(4x+8x^2+\frac{32}{3}x^3+\cdots\right)

=-8x^2-\frac{32}{3}x^3+\cdots

And denominator:

4x(e^{4x}-1)=4x\left(4x+8x^2+\cdots\right)=16x^2+32x^3+\cdots

Therefore,

b=\lim_{x\to 0} \frac{-8x^2-\frac{32}{3}x^3+\cdots}{16x^2+32x^3+\cdots}=-\frac{8}{16}=-\frac12

Now compute:

a-2b=4-2\left(-\frac12\right)=4+1=5

Hence the required integer is

\boxed{5}

Comparison with stored answer: Stored correct answer = $5$ . Our derived answer also equals $5$ , so they agree.

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