Find $\lim_{x\to 0^+}\sin(x)\ln(x)$

Question

By using l'Hôpital rule: because we will get $0\times\infty$ when we substitute, I rewrote it as: $$\lim_{x\to0^+}\dfrac{\sin(x)}{\dfrac1{\ln(x)}}$$ to get the form $\dfrac 00$

Then I differentiated the numerator and denominator and I got: $$\dfrac{\cos x}{\dfrac{-1}{x(\ln x)^2}}$$

when substitute in this form I get: $\dfrac{1}{0\times\infty^2}$

Can we have the result $0\times\infty^2=0$? Then the limit will be $\dfrac10=\infty$?

robjohn · Accepted Answer · 2014-12-19T23:06:18.910

8

Hint: do you know how to compute $$ \lim_{x\to0}\frac{\sin(x)}{x} $$ and $$ \lim_{x\to0}x\log(x)=\lim_{x\to0}\frac{\log(x)}{1/x} $$ If so, then you can use $$ \lim_{x\to0}f(x)g(x)=\lim_{x\to0}f(x)\lim_{x\to0}g(x) $$ provided the limits on the right hand side exist.

edited Dec 19 '14 at 23:06

answered Dec 19 '14 at 22:58

robjohn

345,667

the first yes, the second i think also yes? – Maher Dec 19 '14 at 23:01
Then $\sin x \ln x$ as same limit as $x\ln x$. (also, I really wonder why the first move of many seems to use L'Hopital, while much simpler and effective methods like this one work better.) – Clement C. Dec 19 '14 at 23:02
@Maher: Try it and then multiply the limits. – robjohn Dec 19 '14 at 23:02
got it. $0$ right? – Maher Dec 19 '14 at 23:06
1

Yes, indeed (it goes to $0$). – Clement C. Dec 19 '14 at 23:07
1

@Maher: I was in the middle of typing "Indeed!" and I was interrupted. However, as Clement says, that is correct. – robjohn Dec 19 '14 at 23:18
it's ok !.. Thanks :) – Maher Dec 19 '14 at 23:20

score 3 · Answer 2 · answered Dec 19 '14 at 23:04

As a rule of thumb, you should keep the logarithm at the numerator: $$ \lim_{x\to0^+}\frac{\log x}{1/\sin x} $$ This is of the form $\infty/\infty$, so we can apply l'Hôpital's theorem: $$ \lim_{x\to0^+}\frac{1/x}{-\cos x/\sin^2x}= \lim_{x\to0^+}-\frac{\sin^2x}{x\cos x} $$ that you should be able to manage.

score 1 · Answer 3 · answered Dec 20 '14 at 00:37

Your computation can continue.

After you have applied L'H rule once, it suffices to compute \begin{align*} \lim x \ln ^2 x &=\lim \frac{\ln ^2 x}{\frac{1}{x}} \\ &=\lim \frac{2 \ln x \cdot \frac{1}{x}}{-\frac{1}{x^2}} \\ &=\lim \frac{2\ln x}{ - \frac{1}{x}} \\ &=0 \end{align*}

score 0 · Answer 4 · answered Dec 19 '14 at 23:19

0

We can use approximation arguments : when $x$ is small $\sin(x) \approx x$ and any polynomial grows faster than logarithm. Hence $\lim_{x \to 0^+} \sin(x) \ln(x) = \lim_{x \to 0^+} x = 0$

answered Dec 19 '14 at 23:19

chandu1729

3,801

Find $\lim_{x\to 0^+}\sin(x)\ln(x)$

4 Answers4