How Do I solves these types of limits such as $\lim_{x\to\infty}\frac{7x-3}{2x+2}$ I guess the best question I have is how does my strategy change when I get a limit such as 
$$\lim_{x\to\infty}\dfrac{7x-3}{2x+2}$$
What is essential to have as an understanding to solve these problems? Help welcomed.  
 A: hint: $\dfrac{7x-3}{2x+2} = \dfrac{7-\dfrac{3}{x}}{2+\dfrac{2}{x}}$
A: Dividing by the highest power of x
$$\lim\limits_{x\to\infty}\frac{7x-3}{2x+2}= \lim\limits_{x\to\infty}\frac{7-\frac{3}{x}}{2+\frac{2}{x}}= \frac{7-0}{2+0} =\frac72$$
Applying L'Hospital's rule
$$\lim\limits_{x\to\infty}\frac{7x-3}{2x+2}= \lim\limits_{x\to\infty}\frac{\frac{d}{dx}[7x-3]}{\frac{d}{dx}[2x+2]}=\frac72 $$
Using L'Hospital's rule for this limit is overkill, so I'd recommend against it in this case.
A: The method you should be taking is to take advantage of algebra, and in some cases L'Hospitals rule.
So for this question, we have $\lim_{x \to \infty} $ $ \frac{7x-3}{2x+2} $
We I'm sure you can see as x approached infinity we have the case of infinity/infinity so it is valid to use the rule. So  (7x-3)'=(7) and (2x+2)'=2 so we have $\lim_{x \to \infty} \frac{7}{2}  = 7/2$
Or as eluded to above, you can take advantage of the fact that $\lim_{x \to \infty} 7/x = 0 $
A: The problem of the limit having been nicely explained, if you already know simple approximations, you could do something more to answer two questions : what is the limit and how is it approached ?
For more generality, let me consider the case of $$A=\frac{ax+b}{cx+d}$$ what, as said in answers can write $$A=\frac{a+\frac bx}{c+\frac dx}=\frac{a(1+\frac b{ax})}{c(1+\frac d{cx})}$$ Now, if you know or remember that, for small values of $y$, $\frac 1{1+y}\approx 1-y$, then $$A\approx \frac a c(1+\frac b{ax})(1-\frac d{cx})=\frac{a}{c}+\frac{b c-a d}{c^2 x}$$ So, depending on the sign of $(bc-ad)$, the limit will be approached from above an below.
Hoping this helps.
