Limit of $y=f(x)=\frac{3x-e^{\sin x}}{5+e^{-x}-\cos x}$

Question

I have the following function:

$$ y=f(x)=\frac{3x-e^{\sin x}}{5+e^{-x}-\cos x} $$

Which is the limit when $x \to \pm \infty$. May you explain why?

Answer 1

When $x$ goes to $+\infty$, the numerator goes to $+\infty$ whereas the denominator is bounded (and remain in, say, $[3,7]$). Hence, the function goes to $+\infty$.

When $x$ goes to $-\infty$, the numerator is equivalent to $3x$ whereas the denominator is equivalent to $e^{-x}$. Hence, your function has the same limit as $\frac{3x}{e^{-x}}=3xe^x$ at $-\infty$, that is $0$.