Q. 54 - 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.  

$\underset{x\to 0}{\lim }{x}^{2x}$

Step-by-Step Solution

Answer

$\underset{x\to 0}{\lim }{x}^{2x}=1$

Step 1. Given information

$\underset{x\to 0}{\lim }{x}^{2x}$

Step 2. Taking log on both sides

$$\begin{gathered} \begin{array}{r}\underset{x\to 0^{+}}{lim} \ln x^{2x}=\underset{x\to 0^{+}}{lim} \left(2x\right)\ln \left(x\right)\\ =2\underset{x\to 0^{+}}{lim} \left(\frac{\ln x}{\frac{1}{x}}\right)\\ =2\underset{x\to 0^{+}}{lim} \left(\frac{\frac{1}{x}}{-x^{-2}}\right)\end{array} \\ =-2\underset{x\to 0^{+}}{lim} \left(\frac{1}{x}\cdot \frac{1}{x^{-2}}\right) \end{gathered}$$

Step 3. Calculating the limit

$\begin{array}{r}\underset{x\to 0^{+}}{lim} x^{2x}=e^0\\ =1\end{array}$