2
$\begingroup$

I'm having trouble finding the limit $$ \lim_{x\to ∞}\frac{x+\sin(x)}{2x+\cos(x)} $$

I've started with $$ \lim_{x\to ∞}\frac{\frac{x}{x}+\sin(x)}{\frac{2x}{x}+\cos(x)}= \lim_{x\to ∞}\frac{1+\sin(x)}{2+\cos(x)}=\frac{1}{2}\lim_{x\to ∞}\frac{\sin(x)}{\cos(x)} $$

but I don't know how to go on from here or if I should've started with a different method.

Thank you in advance!

$\endgroup$
1
  • 6
    $\begingroup$ All your identities are wrong. $\endgroup$
    – user65203
    Oct 21, 2017 at 10:04

2 Answers 2

7
$\begingroup$

It should be $$ \lim_{x\to ∞}\frac{x+\sin{x}}{2x+\cos{x}}=\lim_{x\rightarrow\infty}\frac{1+\frac{\sin{x}}{x}}{2+\frac{\cos{x}}{x}}=\frac{1}{2}$$

$\endgroup$
2
$\begingroup$

Alternatively: $$\frac{x-1}{2x+1}\le \frac{x+\sin{x}}{2x+\cos{x}}\le \frac{x+1}{2x-1} \Rightarrow$$ $$\lim_{x\to ∞}\frac{x-1}{2x+1}\le \lim_{x\to ∞}\frac{x+\sin(x)}{2x+\cos(x)}\le \lim_{x\to ∞}\frac{x+1}{2x-1} \Rightarrow$$ $$\lim_{x\to ∞}\frac{x+\sin(x)}{2x+\cos(x)}=\frac12.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .