# Evaluate $\lim_{x\rightarrow -1^+}\frac{{e^{\frac{x}{x+1}}}}{x+1}$

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?

• $(-1)^+$ or $-(1^+)$ ?
– user65203
May 12 '21 at 12:19
• I believe it is $-1+0$.
– Gary
May 12 '21 at 12:20
• Did you try expanding the numerator in a series? May 12 '21 at 12:21
• @PhilFreedenberg The exponent in the numerator is large near $x=-1$. What type of series are you thinking about?
– Gary
May 12 '21 at 12:27

Hint: $$\mathop {\lim }\limits_{x \to - 1^ + } \frac{1}{{x + 1}}e^{\frac{x}{{x + 1}}} = \mathop {\lim }\limits_{x \to - 1^ + } \frac{1}{{x + 1}}e^{ 1- \frac{1}{{x + 1}}} = e\mathop {\lim }\limits_{x \to - 1^ + } \frac{1}{{x + 1}}e^{ - \frac{1}{{x + 1}}} = e\mathop {\lim }\limits_{t \to + \infty } te^{ - t} .$$
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 $$\frac 1t = y\to \infty$$: $$e\cdot ye^{-y}\to 0$$