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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?
user3925803
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2 Answers2

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

Sai Mehta
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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.

Sine of the Time
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user3925803
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  • just let $1/x=t$ and you have the fundamental limit – Sine of the Time Nov 01 '23 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. – Arthur Nov 03 '23 at 05:43