If lim [x->0] [(sqrt(ax+b)-2)/x] = 1, find a, b

T

toebo

New member
Joined
Sep 6, 2007
Messages
3
If the limit as x approaches 0 of ((sqrt(ax+b)-2)/x) is equal to 1, what is the value of a and b! I've tried everything! :cry:
 
Re: I need help solving a limit equation!!!!

toebo said:
If the limit as x approaches 0 of ((sqrt(ax+b)-2)/x) is equal to 1, what is the value of a and b! I've tried everything! :cry:
Click to expand...

Please show us some of your work - so that we know where to start while helping you.

Think L'Hospital's Rule.
 
I tried multiplying the left hand side of the equation by the conjugate of the numerator and I got the limit as x approaches 0 of ((ax + b - 4)/(x(sqrt(ax+b)+2)) is equal to 1. I can't figure out how to get rid of the factor of x in the denominator on the left hand side of the equation.
 
\(\displaystyle \L\\\lim_{x\to\0}\frac{\sqrt{ax+b}-2}{x}\)

I am not so sure we can use L'Hopital. It is not an indeterminate form.

We have \(\displaystyle \frac{\sqrt{b}-2}{0}\), not 0/0.

This appears to be undefined to me.

Am I looking at it wrong?.
 
galactus said:
\(\displaystyle \L\\\lim_{x\to\0}\frac{\sqrt{ax+b}-2}{x}\)

I am not so sure we can use L'Hopital. It is not an indeterminate form.

We have \(\displaystyle \frac{\sqrt{b}-2}{0}\), not 0/0.

This appears to be undefined to me.

Am I looking at it wrong?.
Click to expand...

To change it to a definite value - 'b' must be such that the limit becomes 0/0.

For that to happen, b = 4

Then the function is:

\(\displaystyle \frac{\sqrt{ax + 4}-2}{x}\)

Then applying L'Hospital we have

a/(2) = [sqrt]( 4)

a = 4
 
OK. My bad. We had to find a and b that made the limit 1. Sorry :oops:

We can also use the conjugate instead of L'Hopital.

\(\displaystyle \L\\\frac{\sqrt{4x+4}-2}{x}\cdot\frac{\sqrt{4x+4}+2}{\sqrt{4x+4}+2}\)


Which gives:

\(\displaystyle \L\\\lim_{x\to\0}\frac{4x}{x(\sqrt{4x+4}+2)}\)

Now, it should fall into place nicely.
 
How would you find 'b' (=4) - without L'Hospital?( Or the logic that numerator must become 0 to able to get to 1).

I guess that logic (to get 0/0) is independent of L'Hospital
 
Subhotosh Khan said:
How would you find 'b' (=4) - without L'Hospital?( Or the logic that numerator must become 0 to able to get to 1). I guess that logic (to get 0/0) is independent of L'Hospital
Click to expand...
I certainly worked without using L'Hospital.
This is actually a good teaching problem.
The only way for \(\displaystyle \lim _{x \to 0} \frac{{f(x)}}{x}\) to exist is for \(\displaystyle \lim _{x \to 0} f(x) = 0\).
So in this case, \(\displaystyle \lim _{x \to 0} \sqrt {ax + b} - 2 = 0\quad \Leftrightarrow \quad b = 4\), and the rest follows
 
You must log in or register to reply here.
Top