Evaulate $\lim_{x\to\infty}\frac{3x^2-36x+12}{5x^2+113x-2}$ Question:  Find the limit, $$\lim_{x\to\infty}\frac{3x^2-36x+12}{5x^2+113x-2}$$
The limit should be $\frac{3}{5}$ since when $x$ approaches infinity since $\frac{3*\infty^2}{5*\infty^2}$ and infinity squared cancels out.  Am I correct?
 A: $\displaystyle\frac{3x^2-36x+12}{5x^2+113x-2}=\displaystyle\frac{3-36\frac{1}{x}+12\frac{1}{x^2}}{5+113\frac{1}{x}-2\frac{1}{x^2}}$(Dividing by numerator and denominator by $x^2$ )
As $\lim_{x\to \infty }(3-36\frac{1}{x}+12\frac{1}{x^2})=3$ and $\lim_{x\to \infty }(5+113\frac{1}{x}-2\frac{1}{x^2})=5$ using the fact that $\lim_{x\to \infty } \frac{1}{x}=\lim_{x\to \infty }\frac{1}{x^2}=0$
So $\displaystyle \lim_{x\to \infty }\frac{3x^2-36x+12}{5x^2+113x-2}=\frac{\lim_{x\to \infty }(3-36\frac{1}{x}+12\frac{1}{x^2})}{\lim_{x\to \infty }(5+113\frac{1}{x}-2\frac{1}{x^2})}=\frac{3}{5}$
A: HINT:
$$\frac{3x^2-36x+12}{5x^2+113x-2} =\frac{3-\frac{36}x+\frac{12}{x^2}}{5+\frac{113}x-\frac2{x^2}}$$

Alternatively  putting $\frac1x=y$ as $x\to\infty,y\to0$
$$\lim_{x\to\infty}\frac{3x^2-36x+12}{5x^2+113x-2} =\lim_{y\to0}\frac{3-36y+12y^2}{5+113y-2y^2} $$
A: For functions such as these, the limit tends toward the ratio of the coefficients, therefore, the answer would be $\frac{3}{5}$. Below is a full calculation. 
$$\frac{3x^2-36x+12}{5x^2+113x-2}$$
If we remove an $x^2$, we will have the following
$$\frac{x^2(3-\frac{36}{x}+\frac{12}{x^2})}{x^2(5+\frac{113}{x}-\frac{2}{x^2})}$$
Now, the $x^2$'s cancel out, leaving us with the following
$$\frac{(3-\frac{36}{x}+\frac{12}{x^2})}{(5+\frac{113}{x}-\frac{2}{x^2})}$$
Now we can take the limit of each of the components seperately. Therefore, the $\lim$ of $\frac{36}{x}$ and $\frac{113}{x}$ as $x \rightarrow 0$, will be $0$. Similarly, the limit of $\frac{2}{x^2}$ and $\frac{12}{x^2}$ will be zero.
Now we are left with the limits of $3$ in the numerator and $5$ in the denominator. The limit of a constant is constant. Thus, your answer is $\frac{3}{5}$ 

In the future, if you have questions such as these or questions where the denominator has a higher power or the numerator has a higher power. You can use the following rules: 


*

*If the exponent of the highest term in the numerator matches the
exponent of the highest term in the denominator, the limit (at both
$\infty$ and $-\infty$) is the ratio of the coefficients of the highest
terms.
If the numerator has the highest term, then the fraction is called "top-heavy". If, when you divide the numerator by the
denominator the resulting exponent on the variable is even, then the
limit (at both $\infty$ and $-\infty$) is $\infty$. If it is odd, then the
limit at $\infty$ is $\infty$, and the limit at $-\infty$ is $-\infty$.
If the denominator has the highest term, then the fraction is called "bottom-heavy" and the limit (at both $\infty$ and $-\infty$) is
zero. 
Source: Wikibooks Calculus
A: $$\lim_{x\to\infty}\frac{ 3x^2 - 36x+12}{5x^2+113x-2}$$
To solve this you have  two ways, one very simple is  to apply L'hopitals' rule two times and get 
$$\lim_{x\to\infty}\frac{ 3x^2 - 36x+12}{5x^2+113x-2}=
\lim_{x\to\infty}\frac{ 6x- 36}{10x+113}=\lim_{x\to \infty}=6/10=3/5$$
or you ca divide by $x^2$ as follow
$$\lim_{x\to\infty}\frac{ 3x^2 - 36x+12}{5x^2+113x-2}=
\lim_{x\to\infty}\frac{ \frac{3x^2 - 36x+12}{x^2}}{\frac{5x^2+113x-2}{x^2}}
= \frac{\displaystyle\lim_{x\to\infty} 3 - 36/x+12/x^2}{\displaystyle \lim_{x\to\infty}5+113/x-2/x^2}= \frac{3}{5}$$
