# Is it possible to find this limit without using L'Hopital?

Is it possible to find this limit without using L'Hopital's Rule?

$$\lim_{x \to \infty}\frac{\sin(\frac{2}{x})+\frac{2}{x}}{\sin(\frac{1}{x})}$$

This is a problem that I came across in a Calc 1 HW. I consider myself to be really good at Calc 1, yet I am unsure of the solution. Hence, I find the problem to be relevant to the community. Here is my best try so far:

1. Use the trig double angle formula to change it to: $$\lim_{x \to \infty}\frac{2\sin(\frac{1}{x})\cos(\frac{1}{x})+2(\frac{1}{x})}{\sin(\frac{1}{x})}$$
2. Let $$u=\frac{1}{x}$$ and separate into two fractions: $$\lim_{u \to 0^+}\left[\frac{2\sin(u)\cos(u)}{\sin(u)}+2\frac{u}{\sin(u)}\right]$$
3. Cancel the $$\sin$$ terms in the first fraction and use the known limit of $$\lim_{u \to 0}\frac{u}{\sin(u)}=1$$ on the second fraction: $$2\cdot1+2\cdot1=4$$ The problem for me is - doesn't the proof of $$\lim_{u \to 0}\frac{u}{\sin(u)}=1$$ use L'Hopital? So does my solution really count?
• would you like to use $-x<\sin x < x$? Commented Oct 29, 2023 at 11:29
• $~\sin(2/x) = 2\sin(1/x)\cos(1/x),~$ and $~\displaystyle \lim_{x\to \infty} \frac{1/x}{\sin(1/x)} = \lim_{y\to 0^+}\frac{y}{\sin(y)} = 1.$ Commented Oct 29, 2023 at 11:59
• This is pretty much the way I would do it. Commented Oct 31, 2023 at 16:00
• There are proofs of that limit that don't use L'Hopital's rule. See here. Commented Oct 31, 2023 at 16:03

## 2 Answers

As $$x\rightarrow\infty$$, $$\frac1x$$ and $$\frac2x$$ become very small so we can use the small angle approximation for $$\sin\theta$$. $$\lim_{x\rightarrow\infty}\frac{\sin\left(\frac2x\right)+\frac2x}{\sin\left(\frac1x\right)}=\lim_{x\rightarrow\infty}\frac{\frac2x+\frac2x}{\frac1x}=4$$

• Also, this is not a valid proof. Commented Oct 29, 2023 at 11:24
• @geetha290krm Why isn't it a valid proof? Sure, you can always flesh out more details, and you might not consider the proof complete, but I can't see how it is incorrect. Commented Oct 29, 2023 at 11:26
• I haven't seen this "small angle approximation" technique before. I don't believe it is typically included in a Calc 1 curriculum, so I would not solve it this way. Commented Nov 1, 2023 at 11:08
• @user3925803 it's basically $\lim_{x\to 0}\frac{\sin x}{x}=1$. Commented Nov 1, 2023 at 11:13

The OP has the solution completely correct. You are allowed to use the known limit of: $$\lim_{x\to 0}\frac{\sin(x)}{x}=1$$ There are proofs of this that do not use L'Hopital's Rule.

• just let $1/x=t$ and you have the fundamental limit Commented Nov 1, 2023 at 11:17
• "There are proofs of this that do not use L'Hopital's rule" That's a good thing, as you need to calculate this limit to learn what the derivative of the sine function is in the first place. Commented Nov 3, 2023 at 5:43