Finding the limit of an expression without the use of L'Hopital

Question

I have the following question:

Calculate the $\lim_{x \to 0} \frac{1-\cos(2x)}{x}$

So to begin I rewrote it as: $$\lim_{x \to 0} \frac{2\sin^2(x)}{x}$$

& I'm wondering whether it would be acceptable to use the small-angle approximation for sine here since the limit tends towards 0 so it would fit the description. However if anyone has an alternative method that doesn't require any approximation I'd love to hear it. Thank you.

Just to be clear, are you not allowed to use $\displaystyle\lim_{x\to 0} \frac {\sin x}x = 1$? — player3236, Nov 05 '20 at 15:40
Hi player3236 I re-read the notes just to make sure and it appears that we're able to use this, thank you. — IKnowNothing, Nov 07 '20 at 12:51

score 4 · Answer 1 · answered Nov 05 '20 at 15:40

4

Because $|\sin(x)|\leq |x|$ for all $x \in \mathbb{R}$, you have $$\left|\frac{2\sin^2(x)}{x} \right| \leq 2|\sin(x)|$$

so you can conclude by squeeze theorem.

answered Nov 05 '20 at 15:40

TheSilverDoe

29,720

score 3 · Accepted Answer · answered Nov 05 '20 at 15:48

Your way is fine and by standard limit we can conclude indeed

$$\frac{2\sin^2(x)}{x}= 2\cdot \left(\frac{\sin x}{x}\right)^2 \cdot x \to 2 \cdot 1 \cdot 0 =0$$

as an alternative by standard limit again

$$\frac{1-\cos(2x)}{x}=4x\cdot\frac{1-\cos(2x)}{(2x)^2} \to 0 \cdot \frac1 2=0$$

Finding the limit of an expression without the use of L'Hopital

2 Answers2