Lim x rightarrow 0 frac e-(1+2 x) 1/2 x x is equal to :

Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 9 April, Shift 2 — Question 1

$\operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to :

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

Let

y=(1+2x)^{\frac{1}{2x}}

Take logarithm:

\ln y=\frac{1}{2x}\ln(1+2x)

Using expansion,

\ln(1+2x)=2x-2x^2+\frac{8x^3}{3}+o(x^3)

Hence,

\ln y = \frac{1}{2x}\left(2x-2x^2+\frac{8x^3}{3}\right) = 1-x+\frac{4x^2}{3}+o(x^2)

So,

y = e^{\,1-x+\frac{4x^2}{3}+o(x^2)} = e\cdot e^{-x+\frac{4x^2}{3}+o(x^2)}

Using $e^u=1+u+\frac{u^2}{2}+o(u^2)$ ,

e^{-x+\frac{4x^2}{3}} = 1-x+\frac{11x^2}{6}+o(x^2)

Thus,

(1+2x)^{\frac{1}{2x}} = e\left(1-x+\frac{11x^2}{6}+o(x^2)\right)

Now numerator:

e-(1+2x)^{\frac{1}{2x}} = e-e\left(1-x+\frac{11x^2}{6}\right) = e\left(x-\frac{11x^2}{6}+o(x^2)\right)

Divide by $x$ :

\frac{e-(1+2x)^{\frac{1}{2x}}}{x} = e\left(1-\frac{11x}{6}+o(x)\right)

Taking limit,

\boxed{e}

Answer key and solution verified before publishing.

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