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I have this limit:

$$ \lim_{x\to\infty}\frac{x^3+\cos x+e^{-2x}}{x^2\sqrt{x^2+1}} $$ I tried to solve it by this:

$$ \lim_{x\to\infty}\frac{x^3+\cos x+e^{-2x}}{x^2\sqrt{x^2+1}} = \lim_{x\to\infty}\frac{\frac{x^3}{x^3}+\frac{\cos x}{x^3}+\frac{e^{-2x}}{x^3}}{\frac{x^2\sqrt{x^2+1}}{x^3}} = \frac{0+0+0}{\frac{\sqrt{\infty^2+1}}{\infty}}$$ I do not think that I got it right there... Wolfram also says that the answer is $1$, which this does not seems to be. How do I solve this?

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    $\begingroup$ In your denominator, write $${x^2\over x^2}{\sqrt{x^2+1}\over x}$$ and see what you can do. Also, note that $x^3/x^3$ isn't zero. $\endgroup$ – Gerry Myerson Mar 3 '14 at 8:31
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For sufficiently large $x$

$\displaystyle \frac{x^3-1}{x^2\sqrt{x^2+1}} \leq \frac{x^3+\cos x+e^{-2x}}{x^2\sqrt{x^2+1}} \leq \frac{x^3+2}{x^2\sqrt{x^2+1}}$

Now,

$\displaystyle \frac{2}{x^2\sqrt{x^2+1}} \rightarrow 0$, and $\displaystyle \frac{-1}{x^2\sqrt{x^2+1}} \rightarrow 0$

Moreover,

$\displaystyle \frac{x^3}{x^2\sqrt{x^2+1}}=\frac{x^3}{x^2\sqrt{x^2(1+\frac{1}{x^2})}}=\frac{x^3}{x^3\sqrt{1+\frac{1}{x^2}}}=\frac{1}{1\sqrt{1+\frac{1}{x^2}}} \rightarrow 1$

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    $\begingroup$ $\rightarrow 1$ $\endgroup$ – Alt Mar 3 '14 at 8:39
  • $\begingroup$ How are you allowed to divide by x² in the squareroot? $\endgroup$ – theva Mar 3 '14 at 8:43
  • $\begingroup$ @Aloizio Macedo: Very nice and elegant answer (+1), but can you please explain how do you obtain the first part of the inequality? Is it by series expansion? Cheers $\endgroup$ – Abhimanyu Arora Mar 3 '14 at 9:54
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    $\begingroup$ @Abhimanyu Arora: The first part comes since $\forall x ,e^{-2x}>0, and \cos(x)\geq-1$, whereas the second is because $e^{-2x} \rightarrow 0$ and $cos(x)\leq 1$ $\endgroup$ – Aloizio Macedo Mar 3 '14 at 14:09
  • $\begingroup$ @theva I did not divide by $x^2$. I put $\sqrt{x^2+1}=\sqrt{x^2 (1+\frac{1}{x^2})}$ and then took off the $x^2$ from the square root (notice that there is an extra $x$ factor outside after I do this. I will edit and make an intermediary step. $\endgroup$ – Aloizio Macedo Mar 3 '14 at 14:12
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A hint: When $x\to\infty$ then numerator and denominator both have order of magnitude $x^3$. Therefore extract a factor $x^3\ne0$ on top and bottom, in the hope that now the numerator and denominator both have a finite limit when $x\to\infty$.

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I think just divide with higher x, like this

(x^4/x^4)*(1/sqrt(1+(x^-2))

so we get , (x^4)/sqrt(x^2+1)

and lim x to->oo we just divide it again with higher x and we get, 1/sqrt(1+0)

and the result 1

sorry ,if i have mistake, i'm just trying my logic

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