Question

a) lim
x --> 1
$\frac{(2 {x} ^{2} - 3x + 1)}{(x - 1)}$
b) lim
x --> -2
$\frac{( {x}^{2} + x - 5) \times (x - 3)}{(x + 3)}$

c) lim
x --> 4/3
$\frac{9 {x}^{2} - 16}{3x + 4}$

d) lim
x-->2
$\frac{ {x}^{3} - 8 }{3 {x}^{2} - 12 }$
e) lim
x-->0
$\frac{ \sqrt{x + 1 - 1} }{x}$
​

Answer 1

a)

$\lim_{x \to 1} \dfrac{(2x^2-3x+1)}{(x-1)}= \lim_{x \to 1} \dfrac{2(x-1)(x-\frac{1}{2})}{(x-1)}=$

$= \lim_{x \to 1} (2x-1)=2(1)-1=\boxed{1}$

b)

$\lim_{x \to -2} \dfrac{(x^2+x-5)(x-3)}{(x+3)}= \dfrac{((-2)^2+(-2)-5)(-2-3)}{(-2+3)}=$

$=\dfrac{(4-7)(-5)}{1}=(-3)(-5)=\boxed{15}$

c)

$\lim_{x \to \frac{4}{3}} \dfrac{9x^2-16}{3x+4}= \lim_{x \to \frac{4}{3}} \dfrac{(3x+4)(3x-4)}{3x+4}=$

$= \lim_{x \to \frac{4}{3}} (3x-4)=3\bigg(\dfrac{4}{3}\bigg)-4=\boxed{0}$

d)

$\lim_{x \to 2} \dfrac{x^3-8}{3x^2-12}= \lim_{x \to 2} \dfrac{(x-2)(x^2+2x+4)}{3(x-2)(x+2)}=$

$= \lim_{x \to 2} \dfrac{x^2+2x+4}{3(x+2)}=\dfrac{2^2+2(2)+4}{3(2+2)}=\dfrac{12}{12}=\boxed{1}$

e)

$\lim_{x \to 0} \dfrac{\sqrt{x+1-1}}{x}= \lim_{x \to 0}\dfrac{\sqrt{x}}{x}=\lim_{x \to 0}\dfrac{1}{\sqrt{x}}=\boxed{\infty}$