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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.

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$f(x)=\frac{1}{3\sqrt{x}}$ is not defined (as real function) if $x<0$. – Hanul Jeon 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
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.).

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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

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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.

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

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

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