Limit at infinity when goes to zero

Question

I have the same limit:

$\displaystyle \lim_{x \to \infty} f(x)= x(a^{1/x}-1) $

At first view the limit of $x$ goes to $\infty$ and the limit of $(a^{1/x}-1)$ is zero because $a^{1/\infty}-1 $$= a^0-1=0 $. Then the product of the limits is zero, but if $a$ is any number, for example, 1000, in my calculator I get the answer $ \ln (1000)$.

Algebraically, how can I show this?

Answer 1

Let $a>0\ \wedge a\ne 1$, then$$\\ \lim_{x\to+\infty}x(a^{\frac{1}{x}}-1)=\lim_{x\to+\infty}\frac{a^{\frac{1}{x}}-1}{\frac{1}{x}}=\\ \\ \\ =\lim_{x\to+\infty}\frac{e^{\frac{\ln(a)}{x}}-1}{\frac{1}{x}}$$. Now we use the substitution $t=\frac{\ln(a)}{x}\implies x=\frac{\ln(a)}{t}$ and when $x\to+\infty, t\to 0$. Using this substitution, the limit becomes $$\lim_{t\to 0}\frac{e^{t}-1}{\frac{t}{\ln(a)}}=\ln(a)\overbrace{\lim_{t\to 0}\frac{e^{t}-1}{t}}^{=1}=\ln(a).$$

Answer 2

You have an indeterminate form, $\infty\cdot 0$. What would it be $0$? It's indeterminate.

To find the limit with only elementary techniques, you can proceed as follows. Let $t= \frac{1}{x}$, so that $t\to 0^+$ as $x\to \infty$. Then, for $x>0$ $$ x(a^{1/x}-1) = \frac{a^{1/x}-1}{1/x} = \frac{a^t-1}{t} = \frac{a^t-a^0}{t-0} $$ so that you should be able to compute
$$ \lim_{x\to\infty} x(a^{1/x}-1) = \lim_{t\to 0^+} \frac{a^t-a^0}{t-0} $$ by recognizing in the second expression the definition of a derivative.

Answer 3

Rephrasing:

Let $e^y=a$ .

Then

$x(e^{y/x}-1)= y\dfrac{e^{y/x}-1}{y/x}$.

Recall: $\lim_{z \rightarrow 0^+} \dfrac{e^z-1}{z}=1$.

Hence:

$\lim_{x \rightarrow \infty} y\dfrac{e^{y/x}-1}{y/x}=y=\log a.$

Note :

$\lim_{z \rightarrow 0^+}\dfrac{e^z-1}{z}= (e^z)'_{z=0}=1$.