Evaluate L = lim as x to 0+ (sin x)^68x

Evaluate $\displaystyle L = \lim_{x\to0^{+}}\bigl(\sin x\bigr)^{68x}$.

Step 1: Take the natural logarithm

As $x\to0^{+}$ the base $\sin x\to0^{+}$ and the exponent $68x\to0$: the form is $0^{0}$. (The right-hand limit is the only one available, since $\sin x\lt 0$ for small negative $x$.) Write $L = e^{I}$ with

\[ I = \lim_{x\to0^{+}}68x\ln\bigl(\sin x\bigr), \]

a product of the form $0\cdot(-\infty)$.

Step 2: Rewrite as a quotient and differentiate

\[ I = 68\lim_{x\to0^{+}}\frac{\ln\bigl(\sin x\bigr)}{1/x}, \]

which is $\dfrac{-\infty}{+\infty}$. With $\dfrac{d}{dx}\ln(\sin x) = \cot x$ and $\dfrac{d}{dx}\left[\dfrac1x\right] = -\dfrac{1}{x^{2}}$, \[I = 68\lim_{x\to0^{+}}\frac{\cot x}{-1/x^{2}} = -68\lim_{x\to0^{+}}x^{2}\cot x = -68\lim_{x\to0^{+}}\frac{x^{2}\cos x}{\sin x}. \tag{1}\]

Step 3: Evaluate the remaining limit

\[ \frac{x^{2}\cos x}{\sin x} = x\cos x\cdot\frac{x}{\sin x} \longrightarrow 0\cdot1 = 0, \tag{2} \]

so $I = -68\cdot0 = 0$.

Step 4: Recover $L$

\[ L = e^{I} = e^{0} = 1. \]

Step 5: Verification

\[\sin x\sim x \;\Longrightarrow\; 68x\ln\sin x\sim68\,x\ln x\to0 \ \text{by Problem 391} \;\;\checkmark\]

\[x = 10^{-4}:\ 0.9392907; \qquad x = 10^{-6}:\ 0.9990610; \qquad x = 10^{-10}:\ 0.9999998 \;\;\checkmark\]

\[\text{The approach is not from nearby: the function dips to } 7.2\cdot10^{-12} \text{ at } x \approx 0.3986 \;\;\checkmark\]