Lazy Limits 3!

$\begin{aligned} L & = \lim_{x \to 0} (1+|\sin x|)^\frac 1x \\ L_+ & = \lim_{x \to 0^+} (1+|\sin x|)^\frac 1x \\ & = \lim_{x \to 0^+} (1+\sin x)^\frac 1x \\ & = \lim_{x \to 0} \left(1+\frac {x\sin x}x\right)^\frac 1x & \small \color{#3D99F6}{\text{Let }x = \frac 1n} \\ & = \lim_{n \to \infty} \left(1+ \frac 1n \cdot \frac {\sin \frac 1n}{\frac 1n}\right)^n \\ & = e \\ L_- & = \lim_{x \to 0^-} (1+|\sin x|)^\frac 1x \\ & = \lim_{x \to 0} (1-\sin x)^\frac 1x \\ & = \frac 1e \end{aligned}$

$\implies L_+ \ne L_-$ , therefore, the limit $\displaystyle L = \lim_{x \to 0} (1+|\sin x|)^\frac 1x$ does not exist.

$\large \displaystyle \lim_{x\to 0} ( 1 + |\sin x|)^{\frac{1}{x}} = e^{\ln \lim_{x\to 0} ( 1 + |\sin x|)^{\frac{1}{x}}} = e^{\lim_{x \to 0} \frac{1}{x} \cdot \ln (1 + |\sin x|)} =$ $\large e^{\frac{|\sin x|}{x}} = \text{ doesn't exist}$ Details. $\lim_{x\to 0^+} \frac{|\sin x|}{x} = 1$ $\lim_{x\to 0^-} \frac{|\sin x|}{x} = -1$

$\lim_{x\to 0} (1+|\sin(x)|)^{\frac{1}{x}}=\lim_{x\to 0} (1+|\sin(x)|)^{\frac{1}{|\sin(x)|}\cdot \frac{|\sin(x)}{x}}$ $=e^{\lim_{x\to 0} |\frac{\sin(x)}{x}|\cdot \frac{|x|}{x}}=e^{\lim_{x\to 0} sgn(x)} =\nexists$