Mathematics

# If $f(x) = \dfrac{2 \sin x- \sin2x}{x^3}$ where $x\neq 0$, then $\lim_\limits {x \to 0} f(x)$ has the value;

$1$

##### SOLUTION

$\lim_{x \to 0}f(x)\\=\lim_{x \to 0}(\frac{2sinx-sin2x}{x^3})\\=\lim_{x \to 0}(\frac{2[x-(\frac{x^3}{3!})+(\frac{x^5}{5!})-....]-[(2x)-(\frac{2x^3}{3!})+(\frac{2x^5}{5!})+....]}{x^3})\\=\lim_{x \to 0}(\frac{x63((\frac{-1}{3})+(\frac{4}{3}))+x^5(....)+higher term}{x^3})\\=(\frac{-1}{3})+(\frac{4}{3})=(\frac{3}{3})=1$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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