Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
use l'Hospital's rule you should see en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule –  Maisam Hedyelloo Jan 28 '13 at 3:40
add comment

5 Answers 5

up vote 11 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, $t = x^2$.


share|improve this answer
    
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
add comment

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 $0$.

share|improve this answer
    
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
add comment

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

share|improve this answer
    
+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
add comment

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 $\lim_{x\to 0}\frac{\sin^2(x^2)}{x^2}=0$.

share|improve this answer
add comment

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

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

share|improve this answer
    
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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.