KVPY 2015 #2

Calculus Level 4

$\large \displaystyle \lim_{x \rightarrow 0} \left( \dfrac{x}{\sin x} \right)^{\frac{6}{x^2}}$

1 solution

Chew-Seong Cheong
Nov 2, 2015

$\begin{aligned} \lim_{x \to 0} \left(\frac{x}{\color{#3D99F6}{\sin x}}\right)^{\frac{6}{x^2}} & = \lim_{x \to 0} \left(\frac{x}{\color{#3D99F6}{x - \frac{x^3}{3!}+\frac{x^5}{5!}-...}}\right)^{\frac{6}{x^2}} \quad \quad \small \color{#3D99F6}{\text{Taylor series}} \\ & = \lim_{x \to 0} \left(\frac{x - \frac{x^3}{6}+\frac{x^5}{120}-...}{x} \right)^{-\frac{6}{x^2}} \\ & = \lim_{x \to 0} \left(1 \color{#3D99F6}{- \frac{x^2}{6}}\right)^{\color{#3D99F6}{-\frac{6}{x^2}}} \quad \quad \quad \quad \quad \quad \space \small \color{#3D99F6}{\text{For }x \to 0 \text{ and let } y = - \frac{x^2}{6}} \\ & = \lim_{x \to 0} (1 {\color{#3D99F6}+y})^{\color{#3D99F6}\frac{1}{y}} = \boxed{e} \end{aligned}$

Where did the rest of the terms of the expansion go in the second last step? Could u pls explain? Thnx

Ayush Garg - 5 years, 5 months ago

When $x \to 0$ , $x^3, x^4, x^5...$ are much smaller than $x^2$ and they are dropped.

Chew-Seong Cheong - 5 years, 5 months ago

Oh..I got it..thnx.

Ayush Garg - 5 years, 4 months ago

KVPY 2015 #2

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