0
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ 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 $x$. $\endgroup$
    – Jam
    Commented Apr 3, 2020 at 9:38

2 Answers 2

2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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. $

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .