JEE Main 2022 — Limits Question with Solution

Given,$\underset{x\to 0}{lim}{\left(\frac{{\left(x+2cosx\right)}^3+2{\left(x+2cosx\right)}^2+3sin\left(x+2cosx\right)}{{\left(x+2\right)}^3+2{\left(x+2\right)}^2+3sin\left(x+2\right)}\right)}^{\frac{100}{x}}$

Now by putting the value of limit we can see it is in form $1^{\infty }$

Now we know that,

$\underset{x\to 0}{\lim }{\left(f\left(x\right)\right)}^{g\left(x\right)}$ when is in form of $1^{\infty }$, we can write this as $e^{\underset{x\to 0}{\lim }\left(f\left(x\right)-1\right)g\left(x\right)}$

Now using the above rule we get,

$=e^{\underset{x\to 0}{\lim }\left[\left(\frac{{\left(x+2\cos x\right)}^3+2{\left(x+2\cos x\right)}^2+3\sin \left(x+2\cos x\right)}{{\left(x+2\right)}^3+2{\left(x+2\right)}^2+3\sin \left(x+2\right)}\right)-1\right]\times \frac{100}{x}}$

$=e^{\underset{x\to 0}{\lim }\left[\frac{100}{x}\left[\frac{{\left(x+2\cos x\right)}^3+2{\left(x+2\cos x\right)}^2+3\sin \left(x+2\cos x\right)-\left({\left(x+2\right)}^3+2{\left(x+2\right)}^2+3\sin \left(x+2\right)\right)}{{\left(x+2\right)}^3+2{\left(x+2\right)}^2+3\sin \left(x+2\right)}\right]\right]}$

$=e^{\underset{x\to 0}{\lim }\frac{100}{x}\left[\left(\frac{{\left(x+2\cos x\right)}^3-{\left(x+2\right)}^3+2{\left(x+2\cos x\right)}^2-2{\left(x+2\right)}^2+3\sin \left(x+2\cos x\right)-3\sin \left(x+2\right)}{8+8+3\sin 2}\right)\right]}$

Using L-hospital method we get,

$=e^{\frac{100}{16+3\sin 2}\underset{x\to 0}{\lim }\frac{3{\left(x+2\cos x\right)}^2\times \left(1+2\sin x\right)-3{\left(x+2\right)}^2+4\left(x+2\cos x\right)\times \left(1-2\sin x\right)-4\left(x+2\right)+3\cos \left(x+2\cos x\right)\times \left(1-2\sin x\right)-3\cos \left(x+2\right)}{1}}$

$=e^{\frac{100}{16+3\sin 2}\left(\frac{12-3\left(4\right)+8\times 1-8+3\cos 2-3\cos 2}{1}\right)}$

$=e^{\frac{100}{16+3\sin 2}\times 0}$

$=e^0=1$