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Question #a4e1a

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Answer 1 · 2017-08-31T22:32:58

As $x \to - \infty$ $xrarr-oo$ , the exponent of $e^{- x}$ $e^-x$ approaches $+ \infty$ $+oo$ , since the minus signs will cancel one another out.

In the denominator, values will approach $- \infty$ $-oo$ but at a much slower rate than which $e^{- x}$ $e^-x$ will approach $+ \infty$ $+oo$ .

The exponential function heavily outweighs just $x$ $x$ , so the function will approach some infinity and not level off at $0$ $0$ .

Note, however, that $e^{- x} > 0$ $e^-x>0$ for all values of $x$ $x$ , and as $x \to - \infty$ $xrarr-oo$ , we see that $x < 0$ $x<0$ . Since the values in the denominator are negative, the overall function will approach $- \infty$ $-oo$ .

Thus, $lim_(xrarr-oo)e^-x/x=-oo$ .

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