Question

Evaluate:
1. ${lim}_{x\to 2}\frac{x^3-3x^2+4}{x^4-8x^2+16}$

2. ${lim}_{x\to 1}[\frac{2}{1-x^2}+\frac{1}{x-1}]$

3. ${lim}_{x\to -3}[\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}]$

Answer 1

1) $\underset{x\to 2}{lim}\frac{x^3-3x^2+4}{x^4-8x^2+16}$

$=\underset{x\to 2}{lim}\frac{(x+1)(x^2-4x+4)}{(x^2-4{)}^2}$

$=\underset{x\to 2}{lim}\frac{(x+1)(x-2{)}^2}{(x-2{)}^2(x+2{)}^2}=\underset{x\to 1}{lim}\frac{x+1}{(x+2{)}^2}$

$=\frac{2}{9}$

2) $\underset{x\to 1}{lim}[\frac{2}{1-x^2}+\frac{1}{x-1}]$

$=\underset{x\to 1}{lim}\frac{(2x-2+1-x^2)}{(1-x^2)(x-1)}$

$=\underset{x\to 1}{lim}\frac{-(x^2-2x+1)}{-(x+1{)}^2(x+1)}=\underset{x\to 1}{lim}\frac{(x-1{)}^2}{(x-1{)}^2(x+1)}$

$=\underset{x\to 1}{lim}\frac{1}{x+1}=\frac{1}{2}$

3) $\underset{x\to -3}{lim}[\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}]$

$=\underset{x\to -3}{lim}\frac{2x^2+12x+18}{(x^2+4x+3)(x^2+8x+15)}$

$=\underset{x\to -3}{lim}\frac{2(x^2+6x+9)}{(x+1)(x+3)(x+3)(x+5)}=\underset{x\to -3}{lim}\frac{2(x+3{)}^2}{(x+1)(x+3{)}^2(x+5)}$

$=\underset{x\to -3}{lim}\frac{2}{(x+1)(x+5)}=\frac{2}{-2\times 2}$

$=\frac{-1}{2}$