# Using the precise definition of limits, show that $\lim\limits_{x\to \infty} \frac{(x+\cos x)}{x}=1$

Use the precise definition of $\lim\limits_{x\to \infty} f(x)=L$ to establish the following limit.

$$\lim_{x\to \infty} \frac{(x+\cos x)}{x}=1$$

What I did is: I assumed $\cos x\ge -1$, then replaced $\frac{x-x}{x}-1 < \epsilon \implies \epsilon \gt -1$. However, I got stuck here since I could not figure out the rest. Explanation would be much appreciated.

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Given $\varepsilon>0$, define $N=1/\varepsilon$.
Then $x \geq N$ implies \begin{align*} \left|\frac{x+\cos x}{x}-1\right| &= \left|\frac{\cos x}{x}\right| \\ &\leq \left|\frac{1}{x}\right| & \text{since } \cos x \leq 1 \text{ for all } x \\ &\leq \frac{1}{N} & \text{since } x \geq N \\ & = \varepsilon. \end{align*}
We conclude that $$\lim_{x \rightarrow \infty} \frac{x+\cos x}{x}=1.$$