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.

Find $$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$$

I know that $\lim_{x \to 0} \frac{\sin x}{x}=1$ But here $\sin \left(\frac{1}{3 \sqrt{x}}\right)$ is given when $x \to 0$. Need help.

share|improve this question
2  
$f(x)=\frac{1}{3\sqrt{x}}$ is not defined (as real function) if $x<0$. –  tetori Jan 2 '13 at 13:01
    
@tetori: $f(x)$ is not defined for $x<0$.So, the limit does not exists.Is this will be the conclution. –  A.D Jan 2 '13 at 13:13

4 Answers 4

up vote 2 down vote accepted

The limit is zero, because $|x| \sin{\frac{1}{3 \sqrt{x}}} \rightarrow 0$ as $x \rightarrow 0$. (The denominator is nonzero in this limit.).

share|improve this answer

The limit of the denominator is $\sqrt7$ so we just need to tame the numerator . Observe $$\left|\left|x\right|\sin\frac{1}{3\sqrt{x}}\right|=\left|x\right|\left|\sin\frac{1}{3\sqrt{x}}\right|\le \left|x\right|\cdot 1$$ What does this tell you? Also note that $x\to 0^+$ as $\sqrt x$ must be defined

share|improve this answer
    
Why you introduce inequality? –  A.D Jan 2 '13 at 13:04
    
@A.D Squeeze. Theorem. –  Nameless Jan 2 '13 at 13:05

You may want to prove the easy and pretty useful

Lemma: If $\,f(x)\xrightarrow [x\to x_0]{}0\,$ and $\,|g(x)|\leq M\,\,\,\forall\,x\in(x_0-\epsilon\,,\,x_0+\epsilon)\,$ , for some $\,\epsilon>0\,$ , then

$$\lim_{x\to x_0}f(x)g(x)=0$$

The above simply says that the limit of a function converging to zero times a bounded function is zero.

Now, since $\,\displaystyle{\left|\sin\frac{1}{3\sqrt x}\right|\leq 1\,\,,\,\,x>0}\,$, we get at once, applying the above lemma to the numerator:

$$\frac{x\sin\frac{1}{3\sqrt x}}{\sqrt{x^4+4x^2+7}}\xrightarrow [x\to 0^+]{}\frac{0}{\sqrt 7}=0$$

Note that as Tetori wrote, your function's defined only for positive $\,x\,$, rendering the absolute value in the numerator useless.

share|improve this answer
    
This is helpful answer.Thanks. –  A.D Jan 2 '13 at 13:40

If we let $1/x=y$ $$\lim_{y \to \infty}\frac{\sin \left(\frac{\sqrt{y}}{3 }\right)}{|y|\sqrt{7}}=0$$

share|improve this answer

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.