# Limit $\lim_{x\to \infty}\frac{3x-\cos(2x)}{4x+\sin x}$

$$\lim_{x\to \infty}\frac{3x-cos2x}{4x+sinx}$$

One way is:

$$\frac{3}{4}=\lim_{x\to \infty}\frac{3x-1}{4x +1}\leq\lim_{x\to \infty}\frac{3x-\cos2x}{4x+\sin x}\leq \lim_{x\to \infty}\frac{3x+1}{4x -1}=\frac{3}{4}$$

Is there a way to solve it using trig identities or l'hopital?

• L'hopital does not apply here. The best way is it think of the comparitive size of x compared to sin(x) or cos(x) in this problem Feb 18, 2020 at 11:42
• @HenryLee, what made you say L'hopital does not apply? since $3x-cos(2x)$, and $4x + sin(x)$ approach infinity, we can apply L'hopital's rule. That is take the derivative of the denominator and the nominator separetly and do the same limit. Feb 18, 2020 at 11:44
• It is useable but just gives the same problem of a trig expression on the top and bottom which is undefined Feb 18, 2020 at 11:45

One easy approach is, $$\lim _{x\to \infty \:}\left(\frac{3-\frac{\cos \left(2x\right)}{x}}{4+\frac{\sin \left(x\right)}{x}}\right)=\frac{\lim _{x\to \infty \:}\left(3-\frac{\cos \left(2x\right)}{x}\right)}{\lim _{x\to \infty \:}\left(4+\frac{\sin \left(x\right)}{x}\right)}=\frac34$$

# $$\lim_{x\rightarrow\infty}\frac{\sin x}{x}=0$$

$$0\le\left|\frac{\sin x}{x}\right|\le\left|\frac{1}{x}\right|\rightarrow0\text{ as }x\rightarrow\infty$$ For any $$\epsilon>0,$$ we find $$\left|\frac{\sin x}{x}\right|<\epsilon$$ for all $$x>\frac{1}{\epsilon}$$

Can you try for $$\lim_{x\rightarrow\infty}\frac{\cos 2x}{x}=0$$

You can indeed use squeezing, but you should do it properly. It is not true that $$\frac{3x-1}{4x+1}\le\frac{3x-\cos2x}{4x+\sin x}$$ However, you can observe that for $$x>1000$$ (one safe bound, it's not relevant which one you choose), both the numerator and the denominator are positive. Moreover, for $$x>1000$$, \begin{align} 0&<3x-1\le 3x-\cos2x\le 3x+1 \\[4px] 0&<4x-1\le 4x+\sin x\le4x+1 \end{align} and therefore $$\frac{3x-1}{4x+1}\le\frac{3x-\cos2x}{4x+\sin x}\le\frac{3x+1}{4x-1}$$ Do you see the quirk? If $$0, then $$0<1/b<1/a$$.

Now you can safely apply squeezing, as $$\lim_{x\to\infty}\frac{3x-1}{4x+1}=\frac{3}{4}=\lim_{x\to\infty}\frac{3x+1}{4x-1}$$ and conclude that also $$\lim_{x\to\infty}\frac{3x-\cos2x}{4x+\sin x}=\frac{3}{4}$$

The simplest method, though is to note that $$\frac{3x-\cos2x}{4x+\sin x}=\frac{3-\dfrac{\cos2x}{x}}{4+\dfrac{\sin x}{x}}$$ and $$\lim_{x\to\infty}\frac{\cos2x}{x}=0=\lim_{x\to\infty}\frac{\sin x}{x}$$

The simplest uses asymptotic equivalents: as sine and cosine are bounded,, we have \begin{aligned}3x-\cos 2x&\sim_\infty 3x,\\4x+\sin x&\sim_\infty 4x, \end{aligned}\biggr\}\quad\text{so} \quad \frac{3x-\cos2x}{4x+\sin x}\sim_\infty \frac{3x}{4x}=\frac3 4.