How to compute $\lim_{x\to \infty} x(a^{\frac{1}{x}}-1)$?

Question

I'm supposed to compute it by using some of the following limits somehow:

$$\lim_{x\to 0}(1+x)^{\frac{1}{x}}=\lim_{x\to \infty}\left(1+\frac{1}{x}\right)^{x}$$

But I can't find how to make this limit appear in the calculation. Can someone give me a hint?

Answer 1

I just thought of writing an answer which would use only the limits you had mentioned:

The given limit, as has been mentioned in my comment and also later in the answer posted above, is equivalent to $$\lim\limits_{y\to 0^+} \frac{a^y-1}{y}.$$

I am just reiterating how this can be evaluated from first principles.

Now, $a^y = e^{y\ln (a)}$. Thus the limit turns out to be $$ \ln(a)\lim\limits_{y\to 0^+} \frac{e^{y\ln(a)} - 1}{y\ln(a)}.$$ Putting $h= y\ln(a)$, the above limit is simply,

$$\ln(a) \lim\limits_{h\to 0^+} \frac{e^h-1}{h}.$$

The limit above, is easily found out from first principles, which you might be knowing. Yet for the sake of completion, I will rewrite it below:

$\textbf{To find $\lim\limits_{h\to 0^+}\frac{e^h-1}{h}$}$.

Put $e^h -1 = t$. Then, $e^h = 1+t$ and $h = \ln (1+t)$. Thus, the limit turn out to be

$$\lim\limits_{t\to 0^+} \frac{1}{\ln(1+t)^{\frac{1}{t}}} = \frac{1}{\ln(e)},$$ where it is required to assume that limit can be taken inside a continuous function.

Well, this does use one of the limits you had mentioned.

Answer 2

The limit you are trying to compute is equivalent to

$$\lim_{t \rightarrow 0} \frac{a^t - 1}{t}$$

because as $x \rightarrow \infty$, we have $t \rightarrow 0$, and this is a basic technique of substitution, and you can find this in every 1st year Calculus book.

Then by $\textbf{L'Hôpital's rule}$, which again you can find in every 1st year Calculus book, we have

$$\lim_{t \rightarrow 0} \frac{a^t - 1}{t} = \lim_{t \rightarrow 0} \frac{(a^t - 1)'}{t'} = \lim_{t \rightarrow 0} \frac{a^t \ln a}{1} = \ln a$$

Therefore, we have

$$\lim_{x \rightarrow \infty} x(a^{\frac{1}{x}} - 1) = \lim_{t \rightarrow 0} \frac{a^t - 1}{t} = \ln a$$