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I have to prove the following equation for homework $$\lim_{x\to \infty }\frac{x^2}{x^2+\sin^2 x}=1$$

The proof must be done by proving that for every $e > 0$ exists a $M > 0$ so that for every $x > M$, $|f(x)-1| < e$ is true.

I can't seem to figure this one out.

I would greatly appropriate anyone who tries to help me out :) Thanks

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Let $r(x)$ represent your fraction. It's not too hard to show that $x^{2} \leq x^{2}+\sin^{2}(x) \leq x^{2}+1$, and so that $1 \geq r(x) \geq x^{2}/(x^{2}+1)$. So, if you can give an epsilon-delta proof that $x^{2}/(x^{2}+1)$ has limit $1$, your result will follow by the squeeze theorem. – Chris Leary Nov 18 '11 at 18:08
up vote 5 down vote accepted

$$\left| \frac{{{x}^{2}}}{{{x}^{2}}+{{\sin }^{2}}x}-1 \right| = \frac{{{\sin(x)}^{2}}}{{{x}^{2}}+{{\sin }^{2}}x} \leq \frac{1}{x^2}$$

Now making $\frac{1}{x^2} \leq \epsilon$ gives you the $M$....

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Great thanks! I can up with $M=1/\sqrt{e}$. is this correct? – Jason Nov 18 '11 at 18:29

$x^2/(x^2+\sin^2x)$ is the reciprocal of the expression over the underbrace below: $$ 1 \le \underbrace{\frac{x^2+\sin^2 x}{x^2}} = 1 + \frac{\sin^2 x}{x^2} \le 1 + \frac{1}{x^2}$$ This is less than $1+\varepsilon$ if $x> \dfrac{1}{\sqrt{\varepsilon}}$.

So take $M =1/\sqrt{\varepsilon}$. Then try to prove that if $1<x<1+\varepsilon$, then $1>1/a>1-\varepsilon$. Then you've got it.

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