NIMCET Limit Previous Year Questions (PYQs) – Page 3 of 3
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Showing 23 questions
If $f(x)=\lim _{{x}\rightarrow0}\, \frac{{6}^x-{3}^x-{2}^x+1}{\log _e9(1-\cos x)}$ is a real number then $\lim _{{x}\rightarrow0}\, f(x)$
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Solution
$\lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}$
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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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