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Evaluate $$\lim_{x\rightarrow -1^+}\frac{{e^{\frac{x}{x+1}}}}{x+1}$$ I came up with this limit recently but I think it's harder than I initially thought, unless I'm doing something wrong. I tried applying L'Hospital, but I quickly noticed that in this case it's unsuccessful, as we'll always have the $\frac{0}{0}$ indeterminate form. Then I tried forming an inequality to perform squeeze theorem, and I got this: $$\frac{2x+1}{(x+1)^{2}}\leq \frac{{e^{\frac{x}{x+1}}}}{x+1} < e^{\frac{x}{x+1}}$$ where the lower bound comes from the inequality $e^x\geq x+1$. Unfortunately, the lower bound goes to $-\infty$ as $x$ goes to $-1^+$, while the upper bound goes to $0$, so this turns out to be unsuccessful as well.

Any ideas on how to solve it?

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  • $\begingroup$ $(-1)^+$ or $-(1^+)$ ? $\endgroup$
    – user65203
    May 12 '21 at 12:19
  • $\begingroup$ I believe it is $-1+0$. $\endgroup$
    – Gary
    May 12 '21 at 12:20
  • $\begingroup$ Did you try expanding the numerator in a series? $\endgroup$ May 12 '21 at 12:21
  • $\begingroup$ @PhilFreedenberg The exponent in the numerator is large near $x=-1$. What type of series are you thinking about? $\endgroup$
    – Gary
    May 12 '21 at 12:27
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Hint: $$ \mathop {\lim }\limits_{x \to - 1^ + } \frac{1}{{x + 1}}e^{\frac{x}{{x + 1}}} = \mathop {\lim }\limits_{x \to - 1^ + } \frac{1}{{x + 1}}e^{ 1- \frac{1}{{x + 1}}} = e\mathop {\lim }\limits_{x \to - 1^ + } \frac{1}{{x + 1}}e^{ - \frac{1}{{x + 1}}} = e\mathop {\lim }\limits_{t \to + \infty } te^{ - t} . $$

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  • $\begingroup$ Thank you! I knew I was missing something, I should've thought of substitution. $\endgroup$
    – rich225
    May 12 '21 at 12:18
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Let $t=x+1\to 0^+$. Then the limit is $$\frac{e^{\frac{t-1}{t}}}{t} = \frac{e^{1-\frac 1t}}{t}= e\cdot \frac{e^{-\frac 1t}}{t} $$ Further, let $\frac 1t = y\to \infty$: $$e\cdot ye^{-y}\to 0$$

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