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I would appreciate some help figuring out a way to solve the following limit: $$\lim_{x\rightarrow \infty} \left(x-e^x \right)$$

I know that $e^x$ is much larger than $x$, and therefore the limit will be $-\infty$. What I would like to know is whether or not there is a way to solve this algebraically. Any one able to help? I can't seem to force L'Hopital's Rule here, nor was I able to use the limit definition of $e$.

Thanks, Mada

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    $\begingroup$ It possibly becomes more explicit if you write $x - e^x = -e^x (1-\frac{x}{e^x})$? $\endgroup$ – flavio Dec 12 '13 at 17:37
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    $\begingroup$ Write the limit expression as $x(1-{e^x\over x})$. Show that $e^x/x$ tends to infinity as $x$ tends to infinity. then deduce your limit is... $\endgroup$ – David Mitra Dec 12 '13 at 17:38
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    $\begingroup$ I really hope your teacher isn't asking you this with the hopes that you rearrange it and then use L'Hopital's rule. $\endgroup$ – hunter Dec 12 '13 at 17:42
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You can do some rearranging: $$\lim_{x\rightarrow \infty}x-e^x=\lim_{x\rightarrow \infty}xe^{-x}/e^{-x}+1/e^{-x}=\lim_{x\rightarrow \infty}\frac{x/e^{x}-1}{e^{-x}}=\lim_{x\rightarrow \infty}-e^{x}=-\infty$$ L'hopitals is used implicity here with $\lim_{x\rightarrow \infty}\frac{x}{e^x}$.

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    $\begingroup$ Just one tip, instead of writing the long \rightarrow which produces $\rightarrow$, write only \to instead which produces the same result: $\to$. +1 through. ;-) $\endgroup$ – user93957 Dec 23 '13 at 20:08

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