Question

evaluate:
limit as x approaches negative infinity 3/(e^x-5)

my professor said the answer should be negative infinity but it's not

Answer 1

Is it \(\large \lim_{x \rightarrow -\infty}\frac{3}{e^x - 5}\) or \(\large \lim_{x \rightarrow -\infty}\frac{3}{e^{x - 5}}\) ?

Answer 2

the first one

Answer 3

\[\large \lim_{x \rightarrow -\infty}\frac{3}{e^x - 5} = -\frac 35\]

Answer 4

can you briefly explain how you got that answer, please?

Answer 5

Because \[\large \lim_{x \rightarrow -\infty}e^x = 0\]

Answer 6

But \[\large \lim_{x \rightarrow -\infty}\frac{3}{e^{x - 5}} = \infty \]
That is, if (x-5) is an exponent, then the limit is +infinity.

Answer 7

oh okay, thank youu!