Q. 52 - Calculus

Question

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

$\underset{x\to 0^{+}}{\lim }{\left(x^2+1\right)}^x$.

Step-by-Step Solution

Answer

The exact value of the limit $\underset{x\to 0^{+}}{\lim }{\left(x^2+1\right)}^x$ is, $1.$

Step 1 . Given information

$\underset{x\to 0^{+}}{\lim }{\left(x^2+1\right)}^x$.

Step 2 . Taking logarithm of the limit.

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

Therefore, the value of the limit is given by,

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }{\left(x^2+1\right)}^x=e^0 \\ =1 \end{gathered}$$

The exact value of the limit is, $1.$