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Limits Continuity and Differentiability question

2022 · 26 Jul · Shift 2 · Q25

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Limits Continuity and Differentiability
2022 · 26 Jul · Shift 2 · Q25

Limits Continuity and Differentiability question

2022 · 26 Jul · Shift 2 · Q25

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

Let

\beta=\mathop {\lim }\limits_{x \to 0} \frac{\alpha x-\left(e^{3 x}-1\right)}{\alpha x\left(e^{3 x}-1\right)}

for some

\alpha \in \mathbb{R}

. Then the value of

\alpha+\beta

is :

A
$\frac{14}{5}$
B
$\frac{3}{2}$
C
$\frac{5}{2}$
D
$\frac{7}{2}$

View written solutionFree

Correct answer: C

We are given

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

We need to find $\alpha+\beta$ .

For the limit to be finite, the numerator must vanish to at least the same order as the denominator as $x\to 0$ .

Using the expansion:

e^{3x}-1=3x+\frac{(3x)^2}{2}+\frac{(3x)^3}{6}+\cdots =3x+\frac{9}{2}x^2+\frac{9}{2}x^3+\cdots

So the numerator is

\alpha x-(e^{3x}-1) =\alpha x-\left(3x+\frac{9}{2}x^2+\frac{9}{2}x^3+\cdots\right) = (\alpha-3)x-\frac{9}{2}x^2-\frac{9}{2}x^3-\cdots

The denominator is

\alpha x(e^{3x}-1)=\alpha x\left(3x+\frac{9}{2}x^2+\cdots\right)=3\alpha x^2+O(x^3).

Since the denominator is of order $x^2$ , for the limit to exist finitely we must have the coefficient of $x$ in the numerator equal to $0$ :

\alpha-3=0 \implies \alpha=3.

Now substitute $\alpha=3$ :

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

Again using

e^{3x}-1=3x+\frac{9}{2}x^2+\frac{9}{2}x^3+\cdots,

we get

3x-(e^{3x}-1)= -\frac{9}{2}x^2-\frac{9}{2}x^3+\cdots

and

3x(e^{3x}-1)=3x\left(3x+\frac{9}{2}x^2+\cdots\right)=9x^2+\frac{27}{2}x^3+\cdots

Therefore,

\beta=\lim_{x\to 0}\frac{-\frac{9}{2}x^2+\cdots}{9x^2+\cdots}=-\frac{1}{2}.

Hence,

\alpha+\beta=3-\frac{1}{2}=\frac{5}{2}.

Checking options:

A: $\frac{14}{5}$
B: $\frac{3}{2}$
C: $\frac{5}{2}$ ✅
D: $\frac{7}{2}$

So the correct answer is Option C.

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