### Home > APCALC > Chapter 3 > Lesson 3.1.1 > Problem3-12

3-12.

Without your calculator, evaluate each limit.

1. $\lim\limits _ { h \rightarrow 0 } \frac { ( 2 ( x + h ) - 3 ) - ( 2 x - 3 ) } { h }$

There is either a hole or a vertical asymptote at $h = 0$. If there is a hole, then the $h$ will cancel out.... and the limit will approach a finite value. If there is a vertical asymptote, then the $h$ will not cancel out... and the limit will not exist. Simplify the numerator to see if the limit exists.

$\lim\limits_{h\rightarrow 0}\frac{\left (2\left ( x+h \right ) -3 \right)-\left ( 2x-3 \right )}{h}=\lim\limits_{h\rightarrow 0}\frac{(2x+2h-3)-2x+3}{h}=\lim\limits_{h\rightarrow 0}\frac{2h}{h}=2$

2. $\lim\limits _ { h \rightarrow 0 } \frac { ( ( x + h ) ^ { 2 } + ( x + h ) ) - ( x ^ { 2 } + x ) } { h }$

Refer to the hint in part (a).