# Limit problem

## Homework Statement

If a and b satisfy ##\lim_{x->0}\frac{\sqrt{ax+b}-5}{x} = \frac{1}{2}##, then a+b equals...

A. -15
B. -5
C. 5
D. 15
E. 30

L'hospital

## The Attempt at a Solution

By using L'hospital, I get b=a^2

Then, I got stuck.. Substituting b=a^2 into the limit equation, but I still can't cancel out the x which is the cause of zero denominator..

blue_leaf77
Homework Helper
Since ##x \to 0##, you can write ##\sqrt{ax+b} = \sqrt{b} \sqrt{1+\frac{ax}{b}}## and then expand ##\sqrt{1+\frac{ax}{b}}## into power series. You will be able to determine ##b## first thanks to the presence of ##-5## in the numerator.

• Simon Bridge and terryds
SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

If a and b satisfy ##\lim_{x->0}\frac{\sqrt{ax+b}-5}{x} = \frac{1}{2}##, then a+b equals...

A. -15
B. -5
C. 5
D. 15
E. 30

L'hospital

## The Attempt at a Solution

By using L'hospital, I get b=a^2

Then, I got stuck.. Substituting b=a^2 into the limit equation, but I still can't cancel out the x which is the cause of zero denominator..
What must be true for ##\displaystyle \ \frac{\sqrt{ax+b}-5}{x}\ ## if L'Hôpital's rule can be applied? In particular what must be true of ##\displaystyle \ \lim_{x->0}(\sqrt{ax+b}-5) \ ?##

• Simon Bridge and terryds
What must be true for ##\displaystyle \ \frac{\sqrt{ax+b}-5}{x}\ ## if L'Hôpital's rule can be applied? In particular what must be true of ##\displaystyle \ \lim_{x->0}(\sqrt{ax+b}-5) \ ?##

It must be zero!
So, b equals 25

and a equals 5

a+b = 30..

Thanks a lot!