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Proving $$\lim_{x\to0} \dfrac{\sin x}{x} = 1$$

I know it can be solved with unit circle but it ends up to become same value.

S L
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Hatsune
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2 Answers2

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Easiest way to evaluate the limit is to use L'Hopital's Rule. Since $\frac{\sin 0}{0}$ is indeterminate, we can apply L'Hopital's Rule.

$\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\cos x}{1}=\lim_{x\to 0}\cos x=\cos 0=1$

Sujaan Kunalan
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    This is fine, except that the most common reason for looking at this limit is to help in proving that the derivative of $\sin$ is $\cos$. – Robert Israel Feb 24 '14 at 07:24
  • If you are doing this, just notice directly that the limit is the definition of $\displaystyle\left.\frac{d\sin x}{dx}\right|_{x=0}$. The problem then becomes to explain why sine is differentiable at $0$, with derivative $1$. – Andrés E. Caicedo Feb 24 '14 at 07:24
  • The détour by L'Hopital is odd: if you know what is $f'$ then use $f'$ directly. – Did Feb 24 '14 at 07:50
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L'Hopital's Rule states that $$\lim_{x\to0} \frac{f(x)} {g(x)}=\lim_{x\to0} \frac{f'(x)} {g'(x)}.$$

Therefore, $$\lim_{x\to0} \frac{\sin(x)} {x}=\lim_{x\to0} \frac{\cos(x)} {1}=\cos(0)=1.$$

  • The détour by L'Hopital is odd: if you know what is $f'$ then use $f'$ directly. – Did Feb 24 '14 at 07:50