How do I evaluate the following limits? $$\lim_{x \to + \infty}e^{1-x\over 1+x}$$ $$\lim_{x \to + \infty} {e^{1-x\over 1+x}\over x}.$$ I don't know if I'm allowed to used L'hospital's rule to simplify, mainly because it doesn't traditionally fall into the main indeterminants forms of for L'hopital's rule. So, I'm not confident in continuing to solve the given limit.


Note that $$\lim_{x\rightarrow +\infty}\frac{1-x}{1+x}=-1$$ Therefore $$\lim_{x\rightarrow +\infty}e^{\frac{1-x}{1+x}}=e^{-1}$$ and $$\lim_{x\rightarrow +\infty}\frac{e^{\frac{1-x}{1+x}}}{x}=0$$

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    $\begingroup$ You might want to add that this procedure is correct because $e^x$ is continuous at $-1$. $\endgroup$ – rubik Apr 13 '15 at 19:48

Yes, you can use L'hospital's rule because:

                   limit x -> 1-x/1+x is has the indeterminate form infinite / infinite.  

The reason being that x goes to +ve or -ve infinity.

So, when you use L'hospital's rule you're answer will be e^-1 or 1/e.


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