Compute $\lim_{x\rightarrow\infty} \frac{x^n+15x^{n-1}+3x-1}{x^{n+1}+7}$ for $n > 0$ Compute
\begin{align*}
\lim_{x\rightarrow\infty}\frac{x^{n} + 15x^{n-1}+ 3x - 1}{x^{n+1} + 7}
\end{align*}
for $n > 0$
I can easily compute this with any fixed number, but when it comes to creating a general solution I'm stumped. Any assistance would be greatly appreciated.
 A: The limit is $0$.  In general, for rational functions, if the numerator is a lower degree than the denominator, the end behaviour is $x\to\infty$, $y\to0$.  This is all based on the fundamental limit
$$\lim_{x\to\infty}\frac{1}{x}=0.$$
Therefore,
\begin{align}
\lim_{x\to\infty}\frac{x^n+15x^{n-1}+3x-1}{x^{n+1}+7}&=\lim_{x\to\infty}\frac{x^n+15x^{n-1}+3x-1}{x^{n+1}+7}\color{blue}{\cdot\frac{\frac1{x^{n+1}}}{\frac1{x^{n+1}}}}\\
&=\lim_{x\to\infty}\frac{\frac1x+\frac{15}{x^2}+\frac3{x^{n}}-\frac{1}{x^{n+1}}}{1+\frac7{x^{n+1}}}\\
&=\frac{\lim\limits_{x\to\infty}\frac1x+\lim\limits_{x\to\infty}\frac{15}{x^2}+\lim\limits_{x\to\infty}\frac3{x^{n}}-\lim\limits_{x\to\infty}\frac{1}{x^{n+1}}}{\lim\limits_{x\to\infty}1+\lim\limits_{x\to\infty}\frac7{x^{n+1}}}\\
&=\frac{0+0+0-0}{1+0}\\
&=0
\end{align}
A: To compute the limit of a fraction as $x\to\infty$, one only needs to compute the limit of the leading terms.
$$\lim_{x \to \infty} \frac{x^n + 15x^{n-1} + 3x + 1}{x^{n+1}+7} = \lim_{x \to \infty} \frac{x^n}{x^{n+1}}=\lim_{x \to \infty}\frac{1}{x}=0.$$
