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I have been trying to evaluate $$\lim_{x\to 0}{\frac{\sin^2x}{2x^2}}$$ Finally, I used the L'Hospital's Theorem and I got the answer $1/2$, but I wonder if there is a way to solve this without this.
I also tried using Squeeze Theorem, but my boundaries were approaching different numbers.

ILoveChess
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3 Answers3

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Hint: $\sin^\prime = \cos$, so that $\sin^\prime(0) = \cos 0 = 1.$ Can you make this quantity appear?

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You can write$$\frac{\sin^2 x}{x^2} = \left(\frac{\sin x}{x}\right)^2 = \left(\frac{\sin x-\sin 0}{x-0}\right)^2 \xrightarrow[x\to 0]{} \left(\sin^\prime(0)\right)^2 = 1$$using continuity of the function $t\mapsto t^2$.

Clement C.
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    This is circular reasoning, because the proof of $\sin' x=\cos x$ usually uses the fact that $\lim_{x\to 0}\frac{\sin x}{x}=1$. – user236182 Feb 07 '16 at 20:36
  • In this question there are some proofs of why $\lim_{x\to 0}\frac{\sin x}{x}=1$. – user236182 Feb 07 '16 at 20:38
  • Depends on the proof. Yours, maybe -- some use the fact that $\sin x = \mathrm{Im}(e^{ix})$ and differentiate the complex function; some use the series representation; some use other techniques (e.g., based on $\cos^2 + \sin^2 =1$)... Whether something is "legitimate" or not is really a matter of background. – Clement C. Feb 07 '16 at 20:38
  • @ClementC. i think limit is a more low level concept, so it should'nt be proven using derivatives. please correct me if you think i am wrong – Max Payne Feb 08 '16 at 09:39
  • It may be - that's for the OP to judge... here is a simple method, only requiring knowledge of derivative of sine (which I would deem basic in almost all context not specifying "from first principles"). Whether it is allowed in the context of the OP's question is up to him/her. – Clement C. Feb 08 '16 at 12:09
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Hint: If you want to use the Squeeze Theorem, you can proceed geometrically and observe that for $x\neq 0$ \begin{align}\frac12|\sin x| \le \frac12|x|\le \frac12|\tan x| &\implies 1\le \frac{|x|}{|\sin x|}\le \frac{1}{|\cos x|}\\ &\implies |\cos x|\le \frac{|\sin x|}{|x|}\le 1\\&\implies\cos^2x\le \frac{\sin^2 x}{x^2}\le1\end{align}

Jimmy R.
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  • This geometric interpretation pre-supposes that you have chosen a scale for the angle that already satisfies this inequality resp. has the limit as 1. Remember, geometrically the angle is an equivalence class of ordered pairs of intersecting lines. – Lutz Lehmann Feb 07 '16 at 21:08
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$\displaystyle\lim_{x\to 0}\frac{\sin^2x}{2x^2}=\frac{1}{2}\lim_{x\to 0}\Bigg(\frac{\sin x}{x}\Bigg)^2=\frac{1}{2}$

PinkyWay
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