Find this limit

Question

Find this limit:

$$\lim_{x \to \infty } x\,(1 - k^{1/x})$$

where $0 < k < 1$ is a constant.

Answer 1

Since $k^{1/x}=e^{\ln(k)/x}$, the power series expansion of the exponential function gives

$$x\cdot (1-k^{1/x})=x\cdot \left(-\frac{\ln(k)}{x}-\frac{1}{2}\frac{\ln(k)^2}{x^2}-\cdots\right)= -\ln(k)-\underbrace{\frac{1}{2} \frac{\ln(k)^2}{x} - \cdots}_{\rightarrow 0}\longrightarrow -\ln(k)$$

as $x\rightarrow \infty$.

Answer 2

Use the rule of de l'Hopital:

$$lim_{x \rightarrow \infty} x(1-k^{1/x}) = lim_{x \rightarrow \infty} \frac{1-k^{1/x}}{1/x} = lim_{x \rightarrow \infty} \frac{-log(x)log(k)e^{\frac{1}{x}log(k)}}{log(x)}=-log(k)$$

where we use de l'Hopital in the second step.

Note that the derivative of $$1-k^{1/x}=1-e^{\frac{1}{x}log(k)}$$ is just $$-log(x)log(k)e^{\frac{1}{x}log(k)}=-log(x)log(k)k^{1/x}.$$

For the last step note that $$lim_{x \rightarrow \infty}e^{\frac{1}{x}log(k)}=lim_{x \rightarrow \infty}k^{1/x}=1.$$