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I'm running into problems with this limit: $$\lim_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}$$

I've tried using l'Hospitals rule, however we will alway keep the $\cos(\pi x)$ expression, as well for $\sin(5\pi x)$. Also, terms do not cancel out with $\sin$ and $\pi$ since the product with $5$.

Can anyone give me a hint solving this?

Thanks in advance

Kind regards,

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    $\begingroup$ You recieved several answers to your questions. You should accept the best one, or explain what is still missing. $\endgroup$
    – 5xum
    Dec 12, 2017 at 10:32

4 Answers 4

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Hint: The functions sine and cosine are bounded. Therefore, even though they can oscillate wildly at infinity, they can be controlled.

This implies that in limit operations, we have the following rules of thumb that are correct if used appropriately:

$$1.\lim_{x\to \infty}(0 \times \sin(x)) = 0$$ $$2.\lim_{x\to \infty}(0 \times \cos(x)) = 0$$ $$3.\lim_{x\to \infty}(\frac{\sin(x)}{\infty}) = 0$$ $$4.\lim_{x\to \infty}(\frac{\cos(x)}{\infty}) = 0$$

In all of the above equalities, $0$ in the parentheses is meant to be taken as something infinitesimal. Note that $3.$ and $4.$ can be thought as special cases of $1.$ and $2.$ respectively.

Now factor out $x^2$ in the numerator and $x^4$ under the square root in the denominator and proceed.

If that hint is not enough, hover your mouse over the orange area:

$$\lim_{x \to -\infty}\frac{6x^2+5\cos\pi x}{\sqrt{x^4+5\sin 5\pi x}} = \lim_{x \to -\infty}\frac{x^2\left(6 + \frac{5\cos \pi x}{x^2}\right)}{\sqrt{x^4(1 + \frac{5\sin 5\pi x}{x^4})}}=\lim_{x \to -\infty}\frac{6 + \frac{5\cos \pi x}{x^2}}{\sqrt{1 + \frac{5\sin 5\pi x}{x^4}}}=6$$

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  • $\begingroup$ What do you mean with $\sin(x)/\infty$? $\endgroup$
    – Therkel
    Dec 7, 2017 at 13:32
  • $\begingroup$ @Therkel: I mean that $\sin(x)$ is in the numerator and in the denominator you have some expression that goes to infinity. Then because $\sin(x)$ is bounded, the numerator cannot grow while the denominator grows to infinity and the fraction goes to $0$. $\endgroup$ Dec 7, 2017 at 13:34
  • $\begingroup$ @GuyFSone: I can't understand why you edited my post. Your edit was erroneous too, as it made the hint box disappear. $\endgroup$ Dec 7, 2017 at 15:46
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Easy using Squeeze theorem

$$ 6x^2-5\le6x^2+5\cos{\pi x}\le 6x^2+5$$ and $$ 6x^4-1\le6x^2+\sin{5\pi x}\le 6x^4+1$$

then for $x<-1$ we have

$$\frac{6x^2-5}{\sqrt{x^4+1}}\le \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}} \le \frac{6x^2+5}{\sqrt{x^4-1}}$$

By squeeze theorem we get $$\lim_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}} =6$$

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Writing out like

$$\frac{6x^2+5\cos\pi x}{\sqrt{x^4+5\sin 5\pi x}} = \frac{x^2\left(6 + \frac{5\cos \pi x}{x^2}\right)}{x^2\sqrt{1 + \frac{5\sin 5\pi x}{x^4}}}$$

should do the trick.

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With equivalents:

Since the trigonometric functions are bounded, $6x^2+5\cos \pi x\sim_\infty 6x^2$, $\;x^4+5\sin 5\pi x\sim_\infty x^4$, so $$\frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}\sim_\infty \frac{6x^2}{\sqrt{x^4}}=6.$$

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