Q. 62 - 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.

$\underset{x\to 0}{\lim } {(\sin 3x)}^{2x}$

Step-by-Step Solution

Answer

$\underset{x\to 0}{\lim } {(\sin 3x)}^{2x}=1$

Step 1. Given information

$\underset{x\to 0}{\lim } {(\sin 3x)}^{2x}$

Step 2. Taking log on both sides

$\begin{array}{r}\underset{x\to 0^{+}}{lim} \ln (\sin 3x{)}^{2x}=\underset{x\to 0^{+}}{lim} 2x\ln (\sin 3x)\\ =\underset{x\to 0^{+}}{lim} \frac{2x}{\frac{1}{\ln (\sin 3x)}}\\ =\underset{x\to 0^{+}}{lim} \frac{2}{-\frac{\cos 3x\cdot 3}{(\ln (\sin 3x){)}^2}\cdot \frac{1}{\sin 3x}}\\ =\underset{x\to 0^{+}}{lim} -\frac{2(\ln (\sin 3x){)}^2\sin 3x}{3\cos 3x}\end{array}$

$\begin{array}{r}=-\underset{x\to 0^{+}}{lim} \frac{\sin 3x\cdot 2(\ln (\sin 3x){)}^2}{3\cos 3x}\\ =-\underset{x\to 0^{+}}{lim} \frac{2}{3}\tan 3x(\ln (\sin 3x){)}^2\\ =-\frac{2}{3}\tan 0(\ln (\sin 0){)}^2\\ =0\end{array}$

Step 3. Calculating the limits

$\begin{array}{r}\underset{x\to 0^{+}}{lim} (\sin 3x{)}^{2x}=e^0\\ =1\end{array}$