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

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?

• Your argument for the positive case has the gist despite lacking details and clarity. Your argument for the negative case however is wrong: The denominator is not growing at a faster rate than the numerator for negative $x$.
– Jam
Commented Apr 3, 2020 at 9:38

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

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 $$\rightarrow 0.$$