Question

Okay, I'm attempting to simplify an expression with radicals in it so I can find the limit. I'll type the function below.

Answer 1

\[\lim_{x \rightarrow 0} \sqrt{x+\Delta x} -\sqrt{x}/(\Delta x)\]

Answer 2

How would you work that?

Answer 3

you need to rationalize the numerator

multiply by

\[\frac{\sqrt{x + \Delta x} + \sqrt{x}}{\sqrt{x + \Delta x} + \sqrt{x}}\]

Answer 4

so you end up with

\[\lim_{\Delta x \rightarrow 0} \frac{\sqrt{ x + \Delta x} - \sqrt{x}}{\Delta x} \times \frac{\sqrt{ x + \Delta x} + \sqrt{x}}{\sqrt{x + \Delta x} + \sqrt{x}}\]

this will result in the difference of 2 squares in the numerator

Answer 5

Oh shoot, now it's seems obvious. I just needed to find the conjugate of the whole numerator instead of just one part, yes?

Answer 6

thats correct... you'll end up with

\[\lim_{\Delta x \rightarrow 0} \frac{x + \Delta x - x}{\Delta x(\sqrt{ x + \Delta x} + \sqrt{x}}\]


eliminate the common factor and in the denominator as delta x approaches 0 you'll end up with the solution.

Answer 7

So would I just leave the simplified version ends up being: \[\lim_{x \rightarrow 0} \frac{ 1 }{\sqrt{x + \Delta x} + \sqrt {x} }\]

Answer 8

correct... just collect the like terms and delta x disappears

Answer 9

And then use substitution to replace delta x with 0? (To find the limit?)

Answer 10

well substitute 1st.... then you can collect the like terms...

\[\frac{1}{\sqrt{x} + \sqrt{x}}\]

Answer 11

Yea. So the limit ends up being \[\frac{ 1 }{ 2\sqrt{x} }\] Correct?

Answer 12

thats it.... thats differentiation by 1st principals.

Answer 13

AH! Thank you! I'm going to close the question now.

Answer 14

hope it helped