Finding a&b in a Limit Problem

[acreech]

lim ((ax+b)[sup:k3zjht91]1/2[/sup:k3zjht91]-2)/x=1
x-0

Find numbers a and b


I don't know where to begin with this problem, is there a formula or something that can be used? Thanks for your help.

 

[galactus]

Rationalize the numerator. That gives ax+b-4 in the numerator.

a and b are the same value.

 

[BigGlenntheHeavy]

$\displaystyle Given: \ \lim_{x\to0}\frac{(ax+b)^{1/2}-2}{x} \ = \ 1, \ find \ a \ and \ b.$

$\displaystyle Hence, \ by \ inspection, \ we \ must \ have \ the \ indeterminate \ form \ \frac{0}{0} \ and \ then \ we$

$\displaystyle must \ have \ D_x[(ax+b)^{1/2}-2] \ = \ 1.$

$\displaystyle Ergo, \ (ax+b)^{1/2}-2 \ = \ 0 \ \implies \ (ax+b) \ = \ 4$

$\displaystyle D_x[(ax+b)^{1/2}-2] \ = \ \frac{a}{2(ax+b)^{1/2}} \ = \ 1 \ \implies \ ax+b \ = \ \frac{a^{2}}{4}.$

$\displaystyle \frac{a^2}{4} \ = \ 4 \ \implies \ a^{2} \ = \ 16, \ a \ = \ \pm4$

$\displaystyle Use \ 4, \ then \ (4x+b)^{1/2} \ = \ 2 \ \implies \ 4x+b \ = \ 4, \ 4(0) +b \ = \ 4, \ b \ = \ 4$

$\displaystyle Therefore, \ the \ \lim_{x\to0}\frac{(ax+b)^{1/2}-2}{x} \ = \ 1 \ \implies \lim_{x\to0}\frac{(4x+4)^{1/2}-2}{x} \ = \ 1.$

$\displaystyle Hence, \ a \ = \ b \ = \ 4, \ QED$