_ x 0 x-x+x^3 6 x^5 = - Vidyadip

MCQ

$\lim _{x \rightarrow 0} \frac{\sin x-x+\frac{x^3}{6}}{x^5}=$

✓
$\frac{1}{120}$
B
$-\frac{1}{120}$
C
$\frac{1}{20}$
D
None of these

✓

Answer

Correct option: A.

$\frac{1}{120}$

(A)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\sin x-x+\frac{x^3}{6}}{x^5}=\lim _{x \rightarrow 0} \frac{\cos x-1+\frac{3 x^2}{6}}{5 x^4}$
$=\lim _{x \rightarrow 0} \frac{-\sin x+\frac{6 x}{6}}{20 x^3}=\lim _{x \rightarrow 0} \frac{-\cos x+1}{60 x^2}=\lim _{x \rightarrow 0} \frac{\sin x}{120 x}$
$=\lim _{x \rightarrow 0} \frac{\cos x}{120}=\frac{1}{120}$

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