# Indeterminate form $0^0$ using L'hospitals rule when calculating $\lim_{x\to0^+} x^{\sin(x)}$

Given the question $$\lim_{x\to0^+} x^{\sin(x)}$$

I have deducted so far that this has the indeterminate form $0^0$ so I have taken the natural logarithm of both sides to give me: $$\lim_{x\to0^+} \ln(y) = \lim_{x\to0^+}\sin(x)\ln(x)$$ This still has an indeterminate form so I rearrange it to start using L'hospitals rule: $$\lim_{x\to0^+}\frac{\ln(x)}{\csc(x)}$$ $$=\lim_{x\to0^+}\frac{\frac1x}{-\csc(x)\cot(x)}$$ $$=\lim_{x\to0^+}\frac{-\sin(x)\tan(x)}{x}$$ $$=\lim_{x\to0^+}\frac{-\cos(x)\sec^2(x)}{1} = -1$$ $$\lim_{x\to0^+}x^{\sin(x)} = \frac1e$$

I don't think this is correct, am I doing something wrong, and if so, where?

• You miscomputed the derivative of $\sin (x)\tan(x)$. We have $(f\cdot g)' = f'\cdot g + f\cdot g'$, not $f'\cdot g'$. – Daniel Fischer Jul 7 '15 at 8:10
• oh wow! what a silly mistake. I shall fix this – Panthy Jul 7 '15 at 8:11

$$\lim_{x\to0^+} x^{\sin(x)}$$

$$=\lim_{x\to0^+} \exp(\ln(x)\sin(x))$$

$$=\exp(\lim_{x\to0^+} \ln(x)\sin(x))$$

$$=\exp(\lim_{x\to0^+} \frac{\ln(x)}{1/\sin(x)})$$

Applying L'Hôpital's rule:

$$=\exp(\lim_{x\to0^+} -\frac{\sin(x)^2}{x\cos(x)})$$

$$=\exp(-1\cdot \lim_{x\to0^+} \frac{1}{\cos(x)}\lim_{x\to0^+} \frac{\sin(x)^2}{x})$$

$$=\exp(-\lim_{x\to0^+} \frac{\sin(x)^2}{x})$$

Applying L'Hôpital's rule:

$$=\exp(-\lim_{x\to0^+} 2\cos(x)\sin(x))$$

$$\color{grey}{e^0=1}$$

$$=\boxed {\color{blue}1}$$

• Is my answer below also correct mehtod? – Panthy Jul 7 '15 at 8:23

It's very simple with equivalents: $$\sin x\ln x\sim_0 x\ln x \xrightarrow[x\to 0^+]{} 0\qquad\text{(basic result)}$$ hence $\;x^{\sin x}=\mathrm e^{\sin x\ln x} \to 1$.

$$\lim_{x\to0^+} \ln(y) = \lim_{x\to0^+}\sin(x)\ln(x)$$ $$\lim_{x\to0^+}\frac{\ln(x)}{\csc(x)}$$ $$=\lim_{x\to0^+}\frac{\frac1x}{-\csc(x)\cot(x)}$$ $$=\lim_{x\to0^+}\frac{-\sin(x)\tan(x)}{x}$$ $$=\lim_{x\to0^+}\frac{-\cos(x)\tan(x)-\sin(x)\sec^2(x)}{1} = \frac01 = 0$$ Therefore $$=\lim_{x\to0^+}x^{\sin(x)} = e^0 = 1$$