Evaluate $\lim \limits_{x \to -1^+} (x+1)^{\frac{1}{x+1}}$

Question

I am trying to evaluate limit $$\lim \limits_{x \to -1^+} (x+1)^{\frac{1}{x+1}}$$ The answer is $0$ but I am not able to get it.

My working:

$$\lim \limits_{x \to -1^+} e^{\frac{ln(x+1)}{x+1}}$$ $$e^{\lim \limits_{x \to -1^+} (\frac{ln(x+1)}{x+1})}$$

$$\Rightarrow \lim \limits_{x \to -1^+} \frac{ln(x+1)}{x+1}$$

Using L'hopital rule:

$$\lim \limits_{x \to -1^+} \frac{1}{x+1}$$

Substituting $-1^+$ into the equation

$$\frac{1}{-1+1} = \infty$$ $$\therefore e^\infty = \infty$$

Am I doing something wrong?

Answer 1

You correctly reduce to evaluating $$ \lim_{x\to -1^+}\frac{\ln(x+1)}{x+1} $$ You cannot apply l'Hôpital here, because the hypotheses are not satisfied. This limit is $-\infty$, because the numerator has limit $-\infty$ and the denominator has limit $0$, but taking positive values.

So your limit is $e^{-\infty}=0$.

You should have realized that something went wrong: for $-1<x<0$, you have $$ \ln(x+1)<0,\qquad x+1>0 $$ so the limit above can't be $\infty$, because the function only takes on negative values in a right neighborhood of $-1$.

Answer 2

Note that the limit can be evaluated directly, as should be always done as first attempt, indeed

$$x+1 \to 0^+\implies \frac1{x+1}\to \infty$$

and $0^\infty$ is not an indeterminate form therefore

$$\lim \limits_{x \to -1^+} (x+1)^{\frac{1}{x+1}}=0^\infty=0$$

As an alternative, to avoid confusion, let $y=x+1$ with $y\to 0^+$ and conclude in the same way.

Answer 3

In

$$\lim \limits_{x \to -1^+} \frac{\ln(x+1)}{x+1}$$

the numerator tends to $-\infty$ and the denominator to $0^+$, so you are not in the conditions of applicability of L'Hospital.

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(But that limit is $-\infty$ and the initial one is $0$.)

Answer 4

As others have pointed out, the flaw in your approach is that $\ln(x+1)/(x+1)$ is of the form $-\infty/0$, not $0/0$, as $x\to-1^+$, so L'Hopital does not apply. In problems like this, it sometimes helps to replace the one-sided limit at a finite value with a limit going to infinity. That is, think of $x$ tending to $-1$ from the right as $x=-1+1/u$ with $u\to\infty$. Then $x+1=1/u$, so we have

$$\lim_{x\to-1^+}(x+1)^{1/(x+1)}=\lim_{u\to\infty}\left(1\over u\right)^u=\lim_{u\to\infty}{1\over u^u}=0$$