What is the limit of $e^-x$ as $x to oo$ ?

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Answer 1 · 2018-04-11T13:47:45

As $x-> oo$ , $y->0$

$e^-x=1/e^x$

When x grows infinitely large, the denominator grows infinitely large, causing y to approach 0, $y->0$ .

Hence when $x->oo$ , $y->0$

This is clearly visible when you graph the function, showing that is has a horizontal asymptote of $y=0$

graph{e^-x [-10, 10, -5, 5]}

What is the limit of e^-xe^-x as x to oo x to oo ?