Q. 49 - Calculus

Question

Calculate each of the limits in Exercises $49-64$. Some of these limits are made easier by considering the logarithm of then limit first, and some are not.

$\underset{x\to 0^{+}}{\lim }{x}^{\ln x}$.

Step-by-Step Solution

Answer

The exact value of the limit $\underset{x\to 0^{+}}{\lim }{x}^{\ln x}$ is, $\infty$.

Step 1 . Given information

$\underset{x\to 0^{+}}{\lim }{x}^{\ln x}$.

Step 2 . Taking logarithm of the limit.

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\ln \left(x^{\ln x}\right)=\underset{x\to 0^{+}}{\lim }\left(\ln x\right)\ln \left(x\right) \\ =\underset{x\to 0^{+}}{\lim }(\ln x{)}^2 \end{gathered}$$

The limit using L'Hopital's rule is given below:

$\underset{x\to 0^{+}}{\lim }(\ln x{)}^2=\underset{x\to 0^{+}}{\lim }\frac{(\ln x{)}^3}{\ln x}$  [ in the form of $\frac{0}{0}$]

                  $=\underset{x\to 0^{+}}{\lim }\frac{3(\ln x{)}^2·\frac{1}{x}}{\frac{1}{x}}$  [ Using L'Hopital's rule]

                   $$\begin{gathered} =\underset{x\to 0^{+}}{\lim }3(\ln x{)}^2·\frac{1}{x^2} \\ =3\underset{x\to 0^{+}}{\lim }\frac{(\ln x{)}^2}{x^2} \end{gathered}$$

                   $=3\underset{x\to 0^{+}}{\lim }\frac{2\left(\ln x\right)·\frac{1}{x}}{2x}$ [Using L'Hopital's rule]
                   $=3\underset{x\to 0^{+}}{\lim }\frac{\left(\ln x\right)}{x^2}$

                   $=3\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{x}}{2x}$  [L'Hopital's rule]

                   $$\begin{gathered} =3\underset{x\to 0^{+}}{\lim }\frac{1}{2x^2} \\ =\frac{3}{2}·\frac{1}{0} \\ =\infty \end{gathered}$$

Step 3 . Therefore, the value of the limit is given by,

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\ln \left(x^{\ln x}\right)=e^{\infty } \\ =\infty \end{gathered}$$

Therefore, the exact value of the limit is, $\infty$.