The Limits at Infinity of e to the power of sinx - x

Question

I want to determine the limit of the function $f(x) = e^{sinx-x}$ as x approaches either positive or negative infinity.

My initial hunch is to break down the function into $e^{sinx} / e^x$. Since the denominator grows at a much faster rate than the numerator, the function approaches 0 as x approaches positive infinity, and approaches positive infinity as x approaches negative infinity. I'm wondering if this is a fair argument?

Answer 1

Your argument is vague. A more precise argument is as follows: $e^{-1-x} \leq f(x) \leq e^{1-x}$. Apply squeeze theorem.

Answer 2

In the fraction

$$\dfrac{e^{\sin x}}{e^x}$$

the numerator is bounded within limits $$(e,\frac{1}{e})$$

So it depends more on the denominator which varies monotonically within limits $$(0, \infty)$$

For denominator $x\rightarrow - \infty$ the fraction $ \rightarrow \infty$

For denominator $x\rightarrow + \infty$ the fraction $ \rightarrow 0. $