Find the value of $\lim_{x\to -\infty}\frac{3^{\sin x}+2x+1}{\sin x-\sqrt{x^2+1}}$
$\lim_{x\to -\infty}\frac{3^{\sin x}+2x+1}{\sin x-\sqrt{x^2+1}}$
Since this is in $\frac{\infty}{\infty}$ form,so i applied L Hospital rule,
$\lim_{x\to -\infty}\frac{3^{\sin x}\cos x\log 3+2}{\cos x-\frac{x}{\sqrt{x^2+1}}}$
But i am stuck here.
Note : In the denominator, when we take $x$ out of the square root, we will get $|x|\sqrt{1+\frac{1}{x^2}}$ Which will become = $-x\sqrt{1+\frac{1}{x^2}}$ (As $x$ is negative)
No need for LH here,
Dividing the numerator and denominator by $x$, the given expression becomes: $$\frac{{3^{\sin x} \over x} + 2 + \frac{1}{x}}{{\sin x \over x} + \sqrt{1+\frac{1}{x^2}}}$$ Now, applying the limit, $$\lim_{x \to -\infty}\frac{{3^{\sin x} \over x} + 2 + \frac{1}{x}}{{\sin x \over x} + \sqrt{1+\frac{1}{x^2}}}=2$$
