Question

What is the lim as h ->0 when f(x)=

Answer 1

\[f(x)= \left(\begin{matrix}\sqrt{(3+h)-\sqrt{3}} \\ h\end{matrix}\right)\]

Answer 2

that's divided over h i dont know why it isnt showing.

Answer 3

\[\lim_{h\to 0} \frac{\sqrt {(3+h)-\sqrt 3}}{h}\]

Answer 4

that's what I intended but the root is over 3+h then - root of 3 all over h

Answer 5

\[\lim_{h\to 0} \frac{\sqrt {(3+h)}-\sqrt 3}{h}\]
You mean this?

Answer 6

Yes. Thanks. That looks better.

Answer 7

I need to rationalize it, right?

Answer 8

Yes :)
Multiply and divide by \(\sqrt {3+h}+ \sqrt 3\)

Answer 9

I get to the point of (3+h-9)/(h(root(3+h)+root(3))) and then I messed up somewhere afterwards

Answer 10

Ok, I'll illustrate
\[\lim_{h\to 0} \frac{\sqrt {(3+h)}-\sqrt 3}{h}\times \frac{\sqrt {3+h}-\sqrt 3}{h}\]
\[\lim_{h \to 0}\frac{(\sqrt {3+h})^2-(\sqrt 3)^2}{h\times (\sqrt{3+h}+\sqrt
3)}\]
\[\lim_{h\to 0} \frac {3+h-3}{h\times (\sqrt {3+h}+\sqrt 3)}\]
Can you take it from here?

Answer 11

Sorry I messed the first step

Answer 12

It's actually this
\[\lim_{h\to 0} \frac{\sqrt {(3+h)}-\sqrt 3}{h}\times \frac{\sqrt {3+h}+\sqrt 3}{\sqrt {3+h}+\sqrt 3}\]

Answer 13

Well, Would the limit then be 0? And I see that I went and multiplied the 3's in the first step LOL. fail.

Answer 14

put h=0, here after simplifying
\[\lim_{h\to 0} \frac {3+h-3}{h\times (\sqrt {3+h}+\sqrt 3)}\]

Answer 15

Wait, I don't know how to simplify that.

Answer 16

\[\lim_{h\to 0} \frac {3+h-3}{h\times (\sqrt {3+h}+\sqrt 3)}\]

\[\lim_{h\to 0} \frac {h}{h \times(\sqrt {3+h}+\sqrt 3}\]
\[\lim_{h\to 0} \frac {h}{h \times(\sqrt {3+h}+\sqrt 3}\]
\[\lim_{h\to 0} \frac {1}{\sqrt {3+h}+\sqrt 3}\]
now put h=0 here

Answer 17

Oh..but what happens when you add roots? (algebra skills=none :/ )

Answer 18

Works just like normal no. s
\[3+3=2\times 3\]
so
\[\sqrt 3+\sqrt 3=2\times \sqrt 3 \ or\ 2\sqrt3\]