1
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$\lim_{x\rightarrow -\infty } \frac{3x^{2}-\sin(5x)}{x^{2}+2}$

A. 0

B. 1

C. 2

D. 3

After using L'hopital's rule again and again I got this expression:

$$\lim_{x\rightarrow -\infty } \frac{6+25\sin(5x)}{2}$$

But how do we proceed, what is the value of $\sin(-\infty)$?

Any help will be appreciated!

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    $\begingroup$ I have no idea how you managed to get a $3$ in the numerator? And $$\lim_{x \to \pm \infty} \sin(x)$$ isn't defined. $\endgroup$ – mattos Apr 24 '17 at 5:35
  • $\begingroup$ Sorry, it's 2. Corrected. $\endgroup$ – onelessproblem Apr 24 '17 at 5:36
  • $\begingroup$ The answer could be 'not defined', but the question asks us to choose from the given options. $\endgroup$ – onelessproblem Apr 24 '17 at 5:37
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    $\begingroup$ I didn't say the answer wasn't defined. The answer is $3$. Split the function as $$\frac{3x^{2}}{x^{2} + 2} - \frac{\sin(5x)}{x^{2} + 2}$$ The first term goes to $3$ and the second goes to $0$ $\endgroup$ – mattos Apr 24 '17 at 5:38
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    $\begingroup$ The function isn't $\sin(5x)$ though. It is $\sin(5x)/(x^{2} + 2)$. The limit for that function is defined. $\endgroup$ – mattos Apr 24 '17 at 5:40
3
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Application of L'hopitals rule for the first time gives,

$$\lim_{x \to -\infty} \frac{6x-5\cos (5x)}{2x}$$

We can try to do it again but it won't be useful $\lim_{x \to -\infty} \sin (x)$ does not exist because it oscillates between $-1$ and $1$ for $\frac{\pi}{2}+2\pi k$ and $\frac{3\pi}{2}+2\pi k$ so L'hopitals rule will not give a sensible answer.

Instead we may try more elementary approaches and write is as follows.

$$\lim_{x \to -\infty} \frac{6-5\frac{\cos(5x)}{x}}{2}$$

At this point note that the quaintly $\frac{\cos (5x)}{x}$ goes to zero because the top is bounded by $-1$ and $1$.

So the answer is $3$.

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  • $\begingroup$ Ok, that makes sense. Thankyou kind stranger. $\endgroup$ – onelessproblem Apr 24 '17 at 5:42
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    $\begingroup$ If we go for the elementary approach then why not divide numerator / denominator of the original expression by $x^2$ and get the limit $3$ directly. There is no need of L'Hospital's Rule here. $\endgroup$ – Paramanand Singh Apr 24 '17 at 7:57
  • $\begingroup$ Definitely but it's good to reaffirm that L'hopitals will work. $\endgroup$ – Ahmed S. Attaalla Apr 24 '17 at 21:08
2
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You can also solve by the Squeeze Theorem $$\lim _{x\to \:-\infty \:}\left(\frac{3x^2-\left(-1\right)}{x^2+2}\right)\le \lim \:_{x\to \:-\infty \:}\left(\frac{3x^2-\sin \left(5x\right)}{x^2+2}\right)\le \lim \:_{x\to \:-\infty \:}\left(\frac{3x^2-1}{x^2+2}\right)$$ $$\color{red}{3}\le \lim \:_{x\to \:-\infty \:}\left(\frac{3x^2-\sin \left(5x\right)}{x^2+2}\right)\le \color{red}{3}$$

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