Question

Find the limits :
$i.$ $\underset{x\to 1}{lim}\left[\frac{x^2+1}{x+100}\right]$
$ii.$ $\underset{x\to 2}{lim}\left[\frac{x^3-4x^2+4x}{x^2-4}\right]$
$iii.$ $\underset{x\to 2}{lim}\left[\frac{x^2-4}{x^3-4x^2+4x}\right]$
$iv.$ $\underset{x\to 2}{lim}\left[\frac{x^3-2x^2}{x^2-5x+6}\right]$
$v.$ $\underset{x\to 1}{lim}[\frac{x-2}{x^2-x}-\frac{1}{x^3-3x^2+2x}]$

Answer 1

$\left(i\right)$ Given : $\underset{x\to 1}{lim}\left[\frac{x^2+1}{x+100}\right]$
Substituting $x=1$
$=\frac{1^2+1}{1+100}$
$=\frac{1+1}{101}$
$=\frac{2}{101}$
$\therefore \underset{x\to 1}{lim}\left[\frac{x^2+1}{x+100}\right]=\frac{2}{101}$

$\left(ii\right)$ Given : $\underset{x\to 2}{lim}\left[\frac{x^3-4x^2+4x}{x^2-4}\right]$
Substituting the given value,
$\underset{x\to 2}{lim}\left[\frac{x^3-4x^2+4x}{x^2-4}\right]=\frac{2^3-4\times 2^2+4\times 2}{2^2-4}$
$=\frac{8-16+8}{4-4}$
$=\frac{0}{0}$
Since it is in $\frac{0}{0}$ form, we have to simplify it.

$\Rightarrow \underset{x\to 2}{lim}\left[\frac{x^3-4x^2+4x}{x^2-4}\right]$
$=\underset{x\to 2}{lim}\frac{x(x^2-4x+4)}{x^2-(2{)}^2}$
$=\underset{x\to 2}{lim}\frac{x(x^2+2^2-2\cdot 2\cdot x)}{(x-2)(x+2)}$
$=\underset{x\to 2}{lim}\frac{x(x-2{)}^2}{(x-2)(x+2)}(\because x\ne 2)$
$=\underset{x\to 2}{lim}\frac{x(x-2)}{x+2}$
Substituing $x=2$
$=\frac{2(2-2)}{2+2}=\frac{2\left(0\right)}{4}=\frac{0}{4}=0$

$\left(iii\right)$ Given : $\underset{x\to 2}{lim}\left[\frac{x^2-4}{x^3-4x^2+4x}\right]$
Substituting the given value,
$\underset{x\to 2}{lim}\left[\frac{x^2-4}{x^3-4x^2+4x}\right]=\frac{2^2-4}{2^3-4\times 2^2+4\times 2}$
$=\frac{4-4}{8-16+8}$
$=\frac{0}{0}$
Since it is in $\frac{0}{0}$ form, we have to simplify it.

$\Rightarrow \underset{x\to 2}{lim}\left[\frac{x^2-4}{x^3-4x^2+4x}\right]$
$=\underset{x\to 2}{lim}\left[\frac{x^2-(2{)}^2}{x(x^2-4x+4)}\right]$
$=\underset{x\to 2}{lim}\left[\frac{(x-2)(x+2)}{x(x^2+2^2-2\cdot 2\cdot x)}\right]$
$=\underset{x\to 2}{lim}\left[\frac{(x-2)(x+2)}{x(x-2{)}^2}\right]$
$=\underset{x\to 2}{lim}\left[\frac{x+2}{x(x-2)}\right]$ $(\because x\ne 2)$
Substituing $x=2$
$=\frac{2+2}{2(2-2)}=\frac{4}{2\left(0\right)}=\frac{4}{0}$
$=\infty$
Which is not defined.

$\left(iv\right)$ Given : $\underset{x\to 2}{lim}\left[\frac{x^3-2x^2}{x^2-5x+6}\right]$
Substituting the given value,
$\underset{x\to 2}{lim}\left[\frac{x^3-2x^2}{x^2-5x+6}\right]=\frac{2^3-2\times 2^2}{2^2-5\times 2+6}$
$=\frac{8-8}{4-10+6}$
$=\frac{0}{0}$
Since it is in $\frac{0}{0}$ form, we have to simplify it.

$\Rightarrow \underset{x\to 2}{lim}\left[\frac{x^3-2x^2}{x^2-5x+6}\right]$
$=\underset{x\to 2}{lim}\frac{x^2(x-2)}{x^2-3x-2x+6}$
$=\underset{x\to 2}{lim}\left(\frac{x^2(x-2)}{x(x-3)-2(x-3)}\right)$
$=\underset{x\to 2}{lim}\left(\frac{x^2(x-2)}{(x-2)(x-3)}\right)$
$=\underset{x\to 2}{lim}\left(\frac{x^2}{x-3}\right)(\because x\ne 2)$
Substituting $x=2$
$=\frac{(2{)}^2}{2-3}$
$=\frac{4}{-1}$
$=-4$
$\therefore \underset{x\to 2}{lim}\left[\frac{x^3-2x^2}{x^2-5x+6}\right]=-4$

$\left(v\right)$ Given : $\underset{x\to 1}{lim}[\frac{x-2}{x^2-x}-\frac{1}{x^3-3x^2+2x}]$
$=\underset{x\to 1}{lim}[\frac{x-2}{x(x-1)}-\frac{1}{x(x^2-3x+2)}]$
$=\underset{x\to 1}{lim}[\frac{x-2}{x(x-1)}-\frac{1}{x\left(x\right(x-2)-1(x-2\left)\right)}]$
$=\underset{x\to 1}{lim}[\frac{x-2}{x(x-1)}-\frac{1}{x(x-1)(x-2)}]$
$=\underset{x\to 1}{lim}\left[\frac{(x-2)(x-2)-1}{x(x-1)(x-2)}\right]$
$=\underset{x\to 1}{lim}\left[\frac{(x-2{)}^2-1^2}{x(x-1)(x-2)}\right]$
$=\underset{x\to 1}{lim}\left[\frac{(x-2-1)(x-2+1)}{x(x-1)(x-2)}\right]$
$=\underset{x\to 1}{lim}\left[\frac{(x-3)(x-1)}{x(x-1)(x-2)}\right]$ $(\because x\ne 1)$
$=\underset{x\to 1}{lim}\left[\frac{x-3}{x(x-2)}\right]$
Substituing $x=1$
$=\frac{1-3}{1(1-2)}$
$=\frac{-2}{1\times -1}$
$=2$
$\therefore \underset{x\to 1}{lim}[\frac{x-2}{x^2-x}-\frac{1}{x^3-3x^2+2x}]=2$