taking the limit of $f(x)$, questions How do I take the limit of the following functions?
I had included some of my thoughts with them.
$\lim_{x\to\infty}\dfrac{4x^3 - 2x + 1}{8x^3 + \sin(x^2) - x^{-1}}$; my thoughts: I am not sure about the bottom since there are the sine function and $-1$ power
$\lim_{x\to\infty}\dfrac{e^x}{x^{x-1}}$; my thoughts: isn't $e^x$ faster since $x^{x-1}$ is one power lower than the top?
$\lim_{x\to\infty}\dfrac{x^x}{x^{x+1}}$; my thoughts: the bottom is faster since is one power higher than the top.
Ty!
 A: Here is the first one
$$ \dfrac{4x^3 - 2x + 1}{8x^3 + \sin(x^2) - x^{-1}}=\dfrac{4x^4 - 2x^2 + x}{8x^4 + x\sin(x^2) - 1}\sim_{\infty} \dfrac{4x^4 }{8x^4 } = \frac{1}{2}$$
Added: Recalling Stirling approximation 

$$ n!=\Gamma(n+1) \sim \left(\frac{n}{e}\right)^n\sqrt{2 \pi n}, $$ 

we have
$$ \frac{e^x}{x^{x-1}}= \frac{x}{x^{x}e^{-x}}\sim_{\infty} \frac{x}{\sqrt{2\pi}\sqrt{x}\Gamma(x+1) }=0. $$
Can you do the other one?
Note: Another approach to do the last two limits is to use the result 

If $\lim_{n\to \infty}\frac{a_{n+1}}{a_{n}} = a$ and $|a|<1$, then $\lim_{n\to \infty}a_{n} =0. $

A: Here are some things to consider:


*

*What does $x^{-1}$ tend to at infinity? How big is $\sin$ compared to a polynomial at infinity?

*Have you tried making everything in terms of $e^x$? What happens when $x$ is larger than $e$... larger than $e^2$, $e^3$?

*Hint: $x^{x+1} = x^x\cdot x$
A: In your first limit, lower powers of $x$ become small by comparison to higher powers as $x\to\infty$, and $x^{-1}\to0$.  And $\sin(x)$ oscillates between $1$ and $-1$, so it becomes negligible by comparison to things like $x^3$, that go to $\infty$.
And "become small" and "becomes negligible" actually means they approach $0$ by comparison, i.e. their ratio to $x^3$ approaches $0$.
Hence you have
$$
\lim_{x\to\infty} \frac{4x^3 + \text{something negligible}}{8x^3 + \text{something negligible}}.
$$
