$\lim_{x\to 0^+} (\sin x)^ {\tan x}$
By Sophia Hammond
I have stumbled on this question in one of my problem sets from Cal I and I'm not sure how to proceed after the last step.
$$\lim_{x\to 0^+} (\sin x)^ {\tan x}$$
// Applying exponential rule $$x=e^{\ln(x)}$$
$$\lim_{x\to 0^+} exp[\,\ln((\sin x)^ {\tan x})\,]$$
// Using natural logarithm property to bring the exponent to the front
$$\lim_{x\to 0^+} exp[\,(\tan x)\ln(\sin x)\,]$$
// Using an algebra trick where : $$x= \frac{1}{\frac{1}{x}}$$
$$\lim_{x\to 0^+} exp\left[\,\frac{\ln(\sin x)}{\frac{1}{\tan x}}\,\right]$$
// After this step, we were taught to check for Hospital's rule, however in this case when you plug the value $0$ in the numerator, you get $\ln(\sin 0)$ which is equal to $\ln(0)$. This is the part that is confusing me since $\ln(0)$ is undefined.
// Could someone please explain to me why this whole limit is equal to $1$? Does it have to do with the fact that $x$ is approaching $0$ from the right side?
$\endgroup$ 11 Answer
$\begingroup$Remember what L'Hospital's rule says: it's not a direct substitution of $x = 0$, it's the limit: what happens as $x$ approaches $0$, and specifically from the right (since we have $0^+$)?
Generally, where the limits exist and all the other nuances associated with the definition, the rule states
$$\lim_{x \to c} \frac{f(x)}{g(x)} =\lim_{x \to c} \frac{f'(x)}{g'(x)}$$
if the original limit is one of the indeterminate forms $0/0$ or $\pm \infty / \pm \infty$.
Now, in your problem, since the exponential function is continuous, we can say
$$\lim_{x\to 0^+} exp\left[\,\frac{\ln(\sin x)}{\frac{1}{\tan x}}\,\right] = exp\left[\,\lim_{x\to 0^+} \left( \frac{\ln(\sin x)}{\frac{1}{\tan x}}\,\right) \right]$$
So notice, as $x \to 0$ from the right - not simply $x=0$ - $\sin(x) \to 0$. Since $\ln(x) \to -\infty$ as $x \to 0$ from the right, and we have $\sin(x) \to 0$, it follows that $\ln(\sin(x))\to -\infty$ as $x \to 0^+$.
Further, as $x \to 0^+$, $\tan(x) \to 0$ and thus $1/\tan(x)\to+\infty$.
Thus, we can apply L'Hospital's rule since the limit as-is is an indeterminate form $\pm \infty / \pm \infty$. Differentiate the numerator and denominator (preferably after swapping $\cot(x) = 1/\tan(x)$ for clarity's sake) and try to find the limit as $x \to 0^+$ after.
$\endgroup$ 1
