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

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?

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  • $\begingroup$ 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 $\endgroup$
    – Henry Lee
    Feb 18, 2020 at 11:42
  • $\begingroup$ @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. $\endgroup$
    – Aven Desta
    Feb 18, 2020 at 11:44
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    $\begingroup$ It is useable but just gives the same problem of a trig expression on the top and bottom which is undefined $\endgroup$
    – Henry Lee
    Feb 18, 2020 at 11:45

3 Answers 3

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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 $\lim_{x\rightarrow\infty}\frac{\cos 2x}{x}=0$

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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 $0<a<b$, 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 $$ \lim_{x\to\infty}\frac{\cos2x}{x}=0=\lim_{x\to\infty}\frac{\sin x}{x} $$

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

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