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How would I solve the following trig equations?

$$\lim_{x\to 0}\frac{\sin^2{x^{2}}}{x^{2}}$$

I am thinking the limit would be zero but I am not sure.

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use l'Hospital's rule you should see – Maisam Hedyelloo Jan 28 '13 at 3:40
up vote 12 down vote accepted

We use $\;\sin^2(x^2) = (\sin x^2)(\sin x^2)$

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

Recall, we're also using the fact that $\lim_{t \to 0} \dfrac{\sin t}{t} = 1$. Here, $t = x^2$.

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I see so basically to solve you factor out sin^2? – Fernando Martinez Jan 28 '13 at 3:54
Yes, exactly: $\sin^2 x^2 = (\sin x^2) \cdot (\sin x^2)$ – amWhy Jan 28 '13 at 3:56
It makes sense thanks – Fernando Martinez Jan 28 '13 at 3:57
I understand it perfectly now. – Fernando Martinez Jan 28 '13 at 4:15

I think you want $$\lim_{x\to 0}\frac{\sin^2 x^2}{x^2}.$$ Rewrite our function as $$\left(\sin x^2\right) \frac{\sin x^2}{x^2}.$$ Now is it easy? Yes, indeed the limit is $0$.

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thanks for the help. – Fernando Martinez Jan 28 '13 at 3:54
You are welcome. I hope that soon you will have some knowledge of LaTeX, so that your questions may be easily understood. – André Nicolas Jan 28 '13 at 3:57

$$\lim_{x\to 0} \frac{ \sin^{2} x^2}{x^2} = \lim_{t\to 0} \frac{\sin^2 t }{t}$$ which is equal to $$ \lim_{t \to 0} \frac{\sin t \sin t \cdot t}{t\cdot t} = \lim _{t\to0} \frac{\sin t}{t} \lim _{t\to0} \frac{\sin t}{t} \lim _{t\to 0} t = 1\cdot 1 \cdot 0 =0$$

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+1, just add a line for let $t=x^2$ and don't bother multiplying by $\frac {t}{t}$, substitution should be more than enough, PS: this answer is the only one making use of substitution nicely. – Arjang Jan 29 '13 at 11:33

Yet another solution, via the Sandwich theorem: We know that $|\sin x|\leq |x|$ for every $x$. Thus, whenever $x\neq 0$ we have $$ 0\leq\left|\frac{\sin^2(x^2)}{x^2}\right| = \frac{\left|\sin(x^2)\right|\cdot\left|\sin(x^2)\right|}{x^2} \leq \frac{x^2\cdot x^2}{x^2}=x^2 . $$ Now since $\lim_{x\to 0} x^2=\lim_{x\to 0}0=0$, the Sandwich rule gives that $\lim_{x\to 0}\frac{\sin^2(x^2)}{x^2}=0$.

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We use $$\sin^2 x =\frac{1-\cos 2x}{2}$$

$$\lim_{x\to 0}\frac{\sin^2 x^2}{x^2}.=\lim_{x\to 0} \frac {1-\cos 2x^2}{2x^2}=0 $$

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This is correct, but:(1) please do not write with giant fonts, and (2) don't overuse colours as this can be distracting. Perhaps you should edit your answer. @Ahmed. Also, how is the right hand limit clearer than the left hand one? – DonAntonio Jan 29 '13 at 11:50
@DonAntonio : Before your comment I submitted an edit, but seems that it has disappeared. I don't understand. – Arjang Jan 29 '13 at 11:55
I rejected it, Arjang. Read the reasons why. – DonAntonio Jan 29 '13 at 11:56
@DonAntonio:Thank you I did not know the Rules of Posts – Ahmed Jan 29 '13 at 12:15
Oh, they are not "rules", @Ahmed: it's just some little advices for the posts to look nicer, that's all. – DonAntonio Jan 29 '13 at 12:17

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