Aspire Faculty ID #18681 · Topic: NIMCET 2026 · Just now
NIMCET 2026

$\lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}$

Solution

$\lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin^3x}$ 
Using expansion: 
$\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}$ 

$x^2+2\cos x-2=x^2+2\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)-2$ 
$=x^2+2-x^2+\frac{x^4}{12}-2$ 

$=\frac{x^4}{12}$ 
Also, 

$\sin x\approx x$ 

So, $x\sin^3x\approx x\cdot x^3=x^4$ 

Therefore, $\lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin^3x}$ 

$=\frac{x^4/12}{x^4}$ $=\frac{1}{12}$

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