What is the horizontal asymptote of y= $3e^(-1/x)$ ?

Question

I plugged in positive and negative infinity into x and got $3e^0$ which gave me a horizontal asymptote of 3, is this correct?

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Answer 1 · 2018-01-23T21:32:15

$y=3$

I agree that $y=3$ is the horizontal asymptote.

Since $-1/x$ approaches 0 as $x$ tends toward both positive and negative infinity, $3e^(-1/x)$ tends toward $3e^0=3$ .