Question

If $x_1$ and $x_2$ are the real and distint roots of $a{x}^2+bx+c=0$ then ${lim}_{x\to x_1}{(1+\sin (a{x}^2+bx+c))}^{\frac{1}{x-x_1}}$ is equal to

Answer 1

We know, $\underset{x\to k}{lt}{[1+f\left(x\right)]}^{g\left(x\right)}$

where $f\left(k\right)=0$ & $g\left(k\right)=\infty$ is equal to $e^{f\left(x\right).g\left(x\right)}$

$\therefore$ $\underset{x\to x_1}{lt}{(1+\sin ({ax}^2+bx+c))}^{\frac{1}{x-x_1}}$

$=\underset{x\to x_1}{lt}{(1+\sin \left[(x-x_1)(x-x_2)\right])}^{\frac{1}{x-x_1}}$

$=e^{\sin \left[(x-x_1)(x-x_2)\right].\frac{1}{x-x_1}}$

$=e^{\underset{x\to x_1}{lt}\sin \left[(x-x_1)(x-x_2)\right]\frac{1}{x-x_1}\frac{x-x_2}{x-x_2}}$

$=e^{(x_1-x_2)}$