<s>[INST] $\lim\limits_{x\to \infty}\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} $ Compute the limit:
$$\lim_{x\to \infty}\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} $$
I worked out this to be $\frac{-3}{8}$. I believe this is correct. I used the sandwich theorem but my issue is that the denominator is sometimes negative which isn't helping me. I tried to use absolute values to get rid of this problem:
$$\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} \leq \frac{3x^3+x^2+1}{5x-8x^3-\frac{\pi}{2}},$$
which I then could show is $\frac{-3}{8}$ using algebra of limits, but I am certain this is wrong.
My other way of thinking is:$$\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} = \frac{3+\frac{1}{x}+\frac{\sin(e^x)}{x^3}}{\frac{5}{x^2}-8+\frac{\arctan(\log x)}{x^3}} 
$$
Note that $-1 \leq \sin(e^x) \leq 1$, so for all $x>0$, we have $\frac{-1}{x^3} \leq \frac{\sin(e^x)}{x^3} \leq \frac{1}{x^3}$, so the limit is $0$.
Note that $\frac{-\pi}{2x^3} \leq \frac{\arctan(\log x)}{x^3} \leq \frac{\pi}{2x^3}$, so the limit is $0$.
Now by taking limits:
$$\frac{3+\frac{1}{x}+\frac{\sin(e^x)}{x^3}}{\frac{5}{x^2}-8+\frac{\arctan(\log x)}{x^3}} =\frac{3+0+0}{0-8+0} = -\frac{3}{8}.$$
I believe my second way of working is correct. Can someone please tell me if I am right ? Also, is there a way of doing with this with absolute values ?
 [/INST] Your method is correct, but you can find the limit  instantly after reading the text reasoning in term of asymptotic behaviour.
$$\lim_{x\to \infty}\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} $$
Note that both $\sin e^x$ and $\tan^{-1}(\log x)$ are bounded, so when $x\to +\infty$ larger powers dominate:
$$\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)}\sim\frac{3x^3}{-8x^3}\to -\frac{3}{8} $$
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