# Math Help - Using Conjugates to Solve Limits

1. ## Using Conjugates to Solve Limits

Here is the problem:

$
\displaystyle \lim_{x\to 0 } \frac{\sqrt{1+h}-1}{h}
$

The limit of the denominator is obviously zero, which means that this equation must be converted to find its limit. This can be done my multiplying by the conjugate of the numerator.

$\displaystyle \lim_{x\to 0 } \frac{\sqrt{1+h}-1}{h}*\frac{\sqrt{1+h}+1}{\sqrt{1+h}+1}$

$
\displaystyle \lim_{x\to 0 } \frac{1+h-1}{h(\sqrt{1+h}+1)}}
$

The h cancels out, leaving:

$
\displaystyle \lim_{x\to 0 } \frac{1}{\sqrt{1+h}+1}}
$

This is equal to:
$
\frac{1}{\displaystyle \lim_{x\to 0 } {\sqrt{1+h}+\displaystyle \lim_{x\to 0 }1}}
$

Substituting $0$ for h, $\sqrt{1+h}$ is equal to 1
$
\frac{1}{\displaystyle \lim_{x\to 0 } 1+\displaystyle \lim_{x\to 0 }1}}
$

$\lim_{x\to 0 } 1 = 1$ Thus, the answer is:

$
\displaystyle \lim_{x\to 0 } \frac{\sqrt{1+h}-1}{h}=\frac{1}{2}
$

2. Correct.

3. Originally Posted by reastland
This is equal to:
$
\frac{1}{\displaystyle \lim_{x\to 0 } {\sqrt{1+h}+\displaystyle \lim_{x\to 0 }1}}
$

Substituting $0$ for h, $\sqrt{1+h}$ is equal to 1
$
\frac{1}{\displaystyle \lim_{x\to 0 } 1+\displaystyle \lim_{x\to 0 }1}}
$

$\lim_{x\to 0 } 1 = 1$ Thus, the answer is:

$
\displaystyle \lim_{x\to 0 } \frac{\sqrt{1+h}-1}{h}=\frac{1}{2}
$
This was all uncalled for. At the step above this, just plug in 0.

Originally Posted by reastland

The h cancels out, leaving:

$
\displaystyle \lim_{x\to 0 } \frac{1}{\sqrt{1+h}+1}}$
$=\displaystyle\frac{1}{\sqrt{1+0}+1}=\frac{1}{2}
$

4. I'm posting this for my study group. I probably should have mentioned that. We have to show every single step, no matter how painful. You're right, though. Those last few steps could have been skipped.

5. Originally Posted by reastland
The limit of the denominator is obviously zero, which means that this equation must be converted to find its limit.
It's because both the numerator and denominator are 0. The form $\frac{0}{0}$ is indeterminate. The form $\frac{a}{0}$ where $a\ne 0$ is not.

Also, all your x's should be h's.

And one last minor point: When you say "the h cancels out" you mean that the two expressions after the limit are equal except when h=0. This is ok by a limit theorem which says that you get the same limit as long as the two functions agree everywhere except possibly at the number you're approaching.