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

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!

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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 $3$.

  • $\begingroup$ Ok, that makes sense. Thankyou kind stranger. $\endgroup$ – onelessproblem Apr 24 '17 at 5:42
  • 1
    $\begingroup$ 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

You can also solve by the Squeeze Theorem $$\lim _{x\to \:-\infty \:}\left(\frac{3x^2-\left(-1\right)}{x^2+2}\right)\le \lim \:_{x\to \:-\infty \:}\left(\frac{3x^2-\sin \left(5x\right)}{x^2+2}\right)\le \lim \:_{x\to \:-\infty \:}\left(\frac{3x^2-1}{x^2+2}\right)$$ $$\color{red}{3}\le \lim \:_{x\to \:-\infty \:}\left(\frac{3x^2-\sin \left(5x\right)}{x^2+2}\right)\le \color{red}{3}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.