$ \lim_{x \to 0^-}xe^{\frac{1}{x}}$?

Question

*I cant use LHopital

I try to calculate: $$\displaystyle \lim_{x \to 0^-}xe^{\frac{1}{x}}$$

i found this: Evaluation of $ \lim_{x\rightarrow \infty}\frac{\ln (x)}{x}$ using Squeeze theorem

But it doesnt help, i cant use series.

Someone gave me an idea like this, but still im stuck:

define t = $e^{\frac{1}{x}} \Rightarrow \ln t = \frac{1}{x} \Rightarrow x = \frac{1}{\ln t}$

Therefore we get:

$$\displaystyle \lim_{x \to 0^-}xe^{\frac{1}{x}} = \displaystyle \lim_{t \to 0}\frac{t}{\ln t}$$

Now what?

Or maybe there is another way?

Let $L$ equal your limit then use $\ln(L)=\dots$; can you finish? — JohnColtraneisJC, Jan 17 '20 at 16:37
See also https://math.stackexchange.com/questions/55468/how-to-prove-that-exponential-grows-faster-than-polynomial — Maximilian Janisch, Jan 17 '20 at 16:37
You should check out the youtube channel: "Black pen red pen". This is just up your street and can be useful for similar work. — Leonhard Euler, Jan 17 '20 at 16:40
As $x \to 0^-$, $x \to 0$ and $e^{\tfrac{1}{x}} \to e^{-\infty} \to 0$. Since both limits exist, you have: $$\lim_{x \to 0^-} xe^{\tfrac{1}{x}} = \left(\lim_{x \to 0^-}x \right)\left(\lim_{x \to 0^-}e^{\tfrac{1}{x}}\right) = 0\cdot 0 = 0$$ — SlipEternal, Jan 17 '20 at 16:41

score 4 · Answer 1 · answered Jan 17 '20 at 16:47

4

You can proceed directly: $$\lim_{x\to 0^-}\left(xe^{\frac 1x}\right)=\lim_{x\to 0^-}x\cdot \lim_{x\to 0^-}e^{\frac 1x}=0\cdot 0=0 $$ Note that $\frac 1x\to -\infty$ as $x\to 0^-$.

answered Jan 17 '20 at 16:47

bjorn93

6,787

score 0 · Answer 2 · answered Jan 17 '20 at 16:40

0

HINT: note that $$ \lim_{x\to 0^-}e^{1/x}=\lim_{y\to -\infty }e^{y} $$

answered Jan 17 '20 at 16:40

Masacroso

30,417

$ \lim_{x \to 0^-}xe^{\frac{1}{x}}$?

2 Answers2