# 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..

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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!