How do I evaluate $\lim _{x\to \infty }\left(x\left(5^{\frac{1}{x}}-1\right)\right)$ without using L'Hopital, and what concept/theory do I use for it?

Question

My teacher gave me this homework after he explained how to do problems such as $\lim _{x\to \infty }\left(\frac{3^{x+2}-7^x}{7^{x-1}+5^{x-1}}\right)$ or $\lim _{x\to \infty }\left(1+\frac{2}{2x+1}\right)^{3x+1}$, but I don't see any resemblance whatsoever between the question and the forms that he explained to me earlier.

Answer 1

Hint: Substitute $y=\frac{1}{x}$ and your limit becomes
$\lim _{y\to 0}\dfrac{5^{y}-1}{y}=\lim _{y\to 0}\dfrac{5^{y}-5^0}{y-0}=f'(0)$, for $f(y)=5^y$.

(Assuming that the problem has not risen from an attempt to calculate the derivative at the certain point. In this case, it would be circular reasoning.)

Also, the examples you've given do not seem to have connection to this.

Answer 2

I assume you are allowed to use

• $\lim_{u\to 0}(1+u)^{\frac{1}{u}} = e$

Now, you may set

• $u = 5^{\frac{1}{x}}-1 \Leftrightarrow x = \frac{\ln 5}{\ln (1+u)}$

and consider $u \to 0^+$:

\begin{eqnarray*} x(5^{\frac{1}{x}}-1) & = & \frac{u}{\ln(1+u)}\cdot \ln 5 \\ & = & \frac{1}{\ln(1+u)^{\frac{1}{u}}}\cdot \ln 5 \\ & \stackrel{u\to 0^+}{\longrightarrow} &\frac{1}{\ln e}\cdot \ln 5 = \ln 5 \end{eqnarray*}

Answer 3

Let $y=\frac{1}{x}$ and note that $5^y=e^{{y\ln5}}$.

Since $e^t=1+t+t^2/2+...$, your expression becomes $\frac{e^{y\ln5}-1}{y}=\ln5+$ terms in $y, y^2,... $

As $y$ tends to $0$ the expression tends to $\ln5$.