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How to prove $\lim_{x\to0}\sin x\log x = 0$

My try:

I tried using expansions of sin and log but that does not seem to work.

user2369284
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4 Answers4

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you can't use any taylor expansion for $\log(x)$

But you can say that $\sin x \sim x$, then you have $$\lim_{x \to 0} x\log x = 0$$

Because $x$ goes to $0$ so much faster than $\log x$ goes to infinity.

Or if you have to justify that, rewrite the limit as

$$\frac{\log(x)}{\frac{1}{x}}$$ then use l'Hopital

Ant
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  • Or with $x=e^{-y}$: $\lim_{y\to\infty} \frac{-y}{e^y}$ – Hagen von Eitzen Feb 23 '14 at 10:32
  • But notice that if you know that $f\sim g$, it is not true that you can always substitute $g$ to $f$ in a limit. – Emanuele Paolini Feb 23 '14 at 10:44
  • @EmanuelePaolini well when you have products it is true. My rule of thumb is that if you substitute and some terms cancel out leaving $0$, then it's wrong and you have to use taylor approximation of higher order :) – Ant Feb 23 '14 at 10:46
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You must know the following fundamentals: $$ \lim_{x \to 0} \frac{\sin x}{x} = 1,\qquad \lim_{x \to 0} x^\alpha \log x = 0. $$

Emanuele Paolini
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you can use L'Hopital twice $$\lim_{x\to0}\sin x\log x =\lim_{x\to0}\frac{\log x}{\csc x} \\=\lim_{x\to0}\frac{\frac{1}{ x}}{-\csc x\cot x} \\=\lim_{x\to0}\frac{\sin^2x}{ x\cos x} \\= \lim_{x\to0}\frac{2\sin x\cos x}{ -x\sin x+\cos x}=0$$

Semsem
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lim x tends to 0 log sin x/1/x =Limx tends to 0cosx/sinx/1/x^(-2) =limx tends to 0 cotx/-1/x^2 =Limx tends to 0 -x^2/tanx =Limx tends to 0 -2x/sec^2x =0

Akanksha singh
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