JEE Main 2022 — Limits Question with Solution

Given,

$\underset{x\to 0}{\lim }\frac{\alpha e^x+\beta e^{-x}+\gamma \sin x}{x{\sin }^{2}x}=\frac{2}{3}$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\alpha e^x+\beta e^{-x}+\gamma \sin x}{x^3{\displaystyle \frac{{\sin }^{2}x}{x^2}}}=\frac{2}{3}$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\alpha e^x+\beta e^{-x}+\gamma \sin x}{x^3}=\frac{2}{3}$

Now using expansion of given function in terms of $x$ we get,

$=\underset{x\to 0}{\lim }\frac{\alpha \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \right)+\beta \left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\dots \right)+\gamma \left(x-\frac{x^3}{3!}+\dots \right)}{x^3}$

Now for limit to exist constant terms should be zero

$\Rightarrow a+\beta =0 ....\left(1\right)$

Also coefficient of $x$ should be zero

$\Rightarrow \alpha -\beta +\gamma =0 .......\left(2\right)$

Also coefficient  of $x^2$ should be zero

$\Rightarrow \frac{\alpha }{2}+\frac{\beta }{2}=0 .......\left(3\right)$

So limit becomes, $\underset{x\to 0}{\lim }\frac{x^3\left(\frac{\alpha }{3!}-\frac{\beta }{3!}-\frac{\gamma }{3!}\right)+x^4\left(\frac{\alpha }{3!}-\frac{\beta }{3!}-\frac{\gamma }{3!}\right)}{x^3}=\frac{2}{3}$

Now putting the value of limit we get,

$\frac{\alpha }{6}-\frac{\beta }{6}-\frac{\gamma }{6}=\frac{2}{3} ........\left(4\right)$

Now on solving equation $1, 2, 3 \& 4$ we get,

$\Rightarrow \alpha =1,\beta =-1,\gamma =-2$

Now putting these in all option we get, $\alpha {\beta }^2+\beta {\gamma }^2+\gamma {\alpha }^2+3=0$