1
$\begingroup$

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?

$\endgroup$
2
  • 2
    $\begingroup$ You miscomputed the derivative of $\sin (x)\tan(x)$. We have $(f\cdot g)' = f'\cdot g + f\cdot g'$, not $f'\cdot g'$. $\endgroup$ – Daniel Fischer Jul 7 '15 at 8:10
  • $\begingroup$ oh wow! what a silly mistake. I shall fix this $\endgroup$ – Panthy Jul 7 '15 at 8:11
3
$\begingroup$

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

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

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

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

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:

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

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

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

$\endgroup$
1
  • $\begingroup$ Is my answer below also correct mehtod? $\endgroup$ – Panthy Jul 7 '15 at 8:23
1
$\begingroup$

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 $.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.