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$$\lim_{x\to0+} (\sin x)^{2/\ln x}$$

Not too sure what method to use. I have tried to follow all the helpful answers and this is my working I have done. Please tell me if I have made a mistake. Many thanks! enter image description here

is this answer correct?

user212645
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  • While it is true that $x^0=1$ for any $x$, it is not true that if $f(x)\to 0$ and $g(x)\to 0$ that $f(x)^{g(x)}=1$. This is what is meant by the sentence "$0^0$ is an indeterminate form." – Thomas Andrews Oct 04 '15 at 14:57

5 Answers5

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$$\lim_{x\to 0^+} x\cot x = \lim_{x\to 0^+} \frac{x}{\sin x}\cdot\cos x = 1, $$ hence by De l'Hopital theorem: $$ \lim_{x\to 0^+}\frac{\log\sin x}{\log x} = \lim_{x\to 0^+} x\cot x = 1, $$ and by exponentiating, then squaring, both sides: $$ \lim_{x\to 0^+}\left(\sin x\right)^{\frac{2}{\log x}} = \color{red}{e^2}.$$

Jack D'Aurizio
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Let $$\displaystyle y = \lim_{x\rightarrow 0^{+}}(\sin x)^{\frac{2}{\ln x}}\;,$$ Now put $x=0+h\;,$ Then $h\rightarrow 0\;,$ Where $h$ is a small positive quantity.

So $$\displaystyle y=\lim_{h\rightarrow 0}(\sin h)^{\frac{2}{\ln h}}\;,$$ Now taking $\ln $ on both side, We get

$$\displaystyle \ln(y) = \lim_{h\rightarrow 0}\frac{2\cdot \ln (\sin h)}{\ln h}\; \left(\frac{\infty}{\infty}\right)$$ form

So we will Use $\bf{D\; L\; Hopital\; Rule}$

So $$\displaystyle \ln(y) = \lim_{h\rightarrow 0} \frac{2\cdot \cos h\cdot h}{\sin h} = 2\Rightarrow \ln(y) = 2\Rightarrow y=e^{2}$$

juantheron
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  • I understand up to the last line to where you say lim h-> 0 2coshh/sinh then I don't understand where you have gotten 2 from. Please help – user212645 Oct 04 '15 at 15:33
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The limit is 0, you put the value in place of x. If the answer results in any real number divided by 0 then it is undefined equation and you'll have to simplify it first before putting the value of limit.

Here you can try to apply the limit directly and if it results in 0/0 form then its not correct, simplify it first and then apply the limit.

Bkt
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A different approach is to make the substitution $x\to e^x$. Then,

$$\begin{align} \lim_{x\to 0^+}\left(\sin x\right)^{2/\log x}=\lim_{x\to -\infty}\left(\sin (e^x)\right)^{2/x} \end{align}$$

Since $$\sin (e^x)=e^x+O(e^{3x})$$

we have $\sin (e^x)^{2/x}=e^2\left(1+O(e^{2x})\right)^{2/x}=e^2\left(1+\frac{2}{x}O(e^{2x})\right)\to e^2\,\,\text{as}\,\,x\to -\infty$

Therefore,

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0^+}\left(\sin x\right)^{2/\log x}=e^2}$$

Mark Viola
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  • Please let me know how I can improve my answer. I really want to give you the best answer I can. If the answer was not useful to warrant an up vote, I am happy to delete. – Mark Viola Oct 21 '15 at 23:12
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Let $f(x) = (\sin x)^{2/\log x}$ and we need to evaluate $\lim_{x \to 0^{+}}f(x)$. If this limit is $L$ then \begin{align} \log L &= \log\left(\lim_{x \to 0^{+}}f(x)\right)\notag\\ &= \lim_{x \to 0^{+}}\log f(x)\text{ (via continuity of log)}\notag\\ &= \lim_{x \to 0^{+}}\frac{2\log \sin x}{\log x}\notag\\ &= 2\lim_{x \to 0^{+}}\dfrac{\log \left(\dfrac{\sin x}{x}\right) + \log x}{\log x}\notag\\ &= 2\left(\lim_{x \to 0^{+}}\dfrac{\log \left(\dfrac{\sin x}{x}\right)}{\log x} + 1\right)\notag\\ &= 2(0 + 1) = 2\notag \end{align} Hence $L = e^{2}$. As with most of the limit problems, LHR / Taylor is not required here.

Paramanand Singh
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