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I have a homework question to find the limit of

$$\underset{x\to \infty }{\mathop{\lim }}\,\frac{2{{x}^{4}}+{{x}^{3}}+{{x}^{2}}\sin x}{{{x}^{2}}-5{{x}^{4}}+{{x}^{3}}\sqrt{x}}$$

I have tried for quite a while but can't seem to figure this one out..

Could someone please help me? Thanks a-lot :)

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1  
You can get the answer by dividing both numerator and denominator by $x^4$. –  Tigran Hakobyan Nov 18 '11 at 11:02
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Picky remark: That's no equation (since there's no "equals" sign in sight!). "Expression" or "function" would be a better word to use in this context. –  Hans Lundmark Nov 18 '11 at 11:10

1 Answer 1

up vote 6 down vote accepted

HINT: When finding the limit of a rational function,

$$ \lim_{x\to \infty} \frac{ P(x) }{Q(x) } $$

we divide through by the highest power that occurs to see the limit.

For example, $$ \lim_{x\to \infty} \frac{ 2x^4 + x^3}{x^2- 5x^4} $$

$$= \lim_{x\to\infty} \frac{ 2+ \frac{1}{x} }{\frac{1}{x^2} - 5 } = \frac{ 2 + 0}{ 0-5} = -\frac{2}{5}.$$

The reason we do this is because the highest power is the fastest increasing term, so all terms which are weaker will disappear in the limit after we divide through. The same principle applies to your limit. What do you think you should divide through by?

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I pretty much got close - but don't understand why $sinx/x^2$ is $0$... $sinx$ is not defined for infinity or am I missing somthing? –  Jason Nov 18 '11 at 11:04
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You are correct - $ \lim_{x\to\infty} \sin x $ does not make sense because it never goes towards a single number, instead it keeps oscillating between $-1$ and $1$ periodically. However, we know it goes only between $-1$ and $1.$ So then, $ \frac{-1}{x^2} \leq\frac{\sin x}{x^2} \leq \frac{1}{x^2}.$ And you know the limits of the left and right sides, so what can you conclude? –  Ragib Zaman Nov 18 '11 at 11:06
    
Thanks that makes it perfectly clear :) –  Jason Nov 18 '11 at 11:10

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