# Verify that L'Hopital's rule is of no help in finding the limit , find the limit, if it exists,by some other methods

Verify that L'Hopital's rule is of no help in finding the limit , the find the limit, if it exists,by some other methods $$a-)\lim_{x\rightarrow +\infty}\frac{x+\sin(2x)}{x}$$ $$b-)\lim_{x\rightarrow +\infty}\frac{x(2+\sin(2x))}{x+1}$$

My work:

a-) It is seen that it is indetermine form of $$\infty / \infty$$ , so i thought to use L'Hopital rule such that $$\lim_{x\rightarrow +\infty}\frac{1+2\cos(2x)}{1}= \infty$$

According to the L'Hopital rule , if we catch finite or $$+\infty,-\infty$$ , it is the answer. So , i thought that the answer is $$\infty$$.However , the answer is $$1$$ .

Why cant I use L'Hopital here , why is the answer $$1$$ instead of $$\infty$$ ?

b-) The answer is "Limit does not exist " , but i do not understand why the limit "does not exist"

Can you help me ?

• For a), $\cos2x$ has no limit, not even $\pm\infty$, as it's bounded but oscillatory.
– J.G.
Commented Aug 3, 2022 at 15:43

In a) note that cosine of whatever argument is between $$-1$$ and $$1$$. So I am not sure how you get $$\infty$$. But since your value varies between $$-1$$ and $$1$$, then the limit of the l'Hopital does not exist. But you can split into $$1+\lim_{x\to\infty}\frac{\sin2x}x=1+0=1$$ The second limit does not exist for the same reason. Divide both numerator and denominator by $$x$$. Then the denominator goes to $$1$$, but the numerator oscillates periodically between $$1$$ and $$3$$, so it does not have a limit.