Limit of $\lim_{x\to\infty}\frac{e^x+\cos x}{e^x-\sin 3x}$ How to find the value of $\lim_{x\to\infty}\frac{e^x+\cos x}{e^x-\sin 3x}$?
Wolfram says it's =1 but I don't know how?
 A: write your Limit in the form $$\frac{1+\frac{\cos(x)}{e^x}}{1-\frac{\sin(3x)}{e^x}}$$
A: I write this answer to show why $\lim_{x\to\infty}\frac{\cos x}{e^x}=0$ in a formal way.
You can divide both denominator and numerator by $e^x$ because $e^x\neq 0$, then we get the equivalent limit (as the other answer shows)
$$\lim_{x\to\infty}\frac{1+\frac{\cos x}{e^x}}{1-\frac{\sin 3x}{e^x}}=\frac{1+0}{1-0}$$
Now, observe that
$$\frac{-1}{e^x}\le\frac{\cos x}{e^x}\le \frac1{e^x},\quad\forall x\in\Bbb R\tag{1}$$
then using the squeeze theorem when we apply limits to (1) we have
$$\lim_{x\to\infty}\frac{-1}{e^x}=0\le\lim_{x\to\infty}\frac{\cos x}{e^x}\le\lim_{x\to\infty}\frac1{e^x}=0\implies\lim_{x\to\infty}\frac{\cos x}{e^x}=0$$
The same rule can be used for the other limit in the denominator.
A: When you see the limit
$$\lim_{x\to\infty}\frac{e^x+\cos x}{e^x-\sin 3x}$$
the first thing that should pop to your mind is that the function $e^x$ grows extremely fast when $x \to \infty$ whereas both $\sin(3x)$ and $\cos(x)$ are limited functions that will never be more than $1$ or less than $-1$. Thus, you may want to find a way to show that indeed the $e^x$ part dominates the other two functions. One such way is to factor $e^x$ out:
$$\lim_{x\to\infty}\frac{e^x+\cos x}{e^x-\sin 3x} = \lim_{x\to\infty} \frac{e^x(1 + \frac{\cos x}{e^x})}{e^x(1 - \frac{\sin(3x)}{e^x})}$$
Obviously the $e^x$ cancel and then you can reason about
$$\lim_{x\to\infty} \frac{1 + \frac{\cos x}{e^x}}{1 - \frac{\sin(3x)}{e^x}}$$
but it is straightforward that both $\frac{\cos x}{e^x}$ and $-\frac{\sin(3x)}{e^x}$ go to $0$ when $x \to \infty$ and thus the original limit is $1$ as it becomes
$$\lim_{x\to\infty} \frac{1 + 0}{1 - 0}$$
