Question

lim (h->0) (sqrt(25+h)-5)/h

I don't really understand this problem.
Can someone help me out?

Answer 1

draw it, please?

Answer 2

@hsjunior if you just put h=0, you'll get the form 0/0 so we need to use limits.

Answer 3

\[\lim_{h\to 0} \frac{\sqrt{25+h}-5}{h}\]
Let's rationalize the numerator. Can you try? @hsjumior

Answer 4

would I have to use l'hopital's theorem?

Answer 5

Not required, multiply and divide by
\[\sqrt{25+h}+5\]

Answer 6

so is it just multiplying by its conjugate?

Answer 7

Yes. multiplying and dividing

Answer 8

does this apply to all problems of this type?

Answer 9

u know L-Hospitals Rule ?

Answer 10

On multiplying and dividing you'd get
\[\lim_{h \to 0}\frac{(\sqrt{25+h})^2-25}{h(\sqrt{25+h}+5)}\]
Now you'll get
\[\lim_{h \to 0}\frac{25+h-25}{h(\sqrt{25+h}+5)}\]
Can you simplify this?

Answer 11

\[\frac{ \sqrt{25+h}-5 }{ h }\times \frac{ \sqrt{25+h}+5 }{ \sqrt{25+5}+5 }\]
So is it like this?

Answer 12

yes :)

Answer 13

and the h then becomes by itself in the numerator, correct?

Answer 14

Recognize this as the limit definition of the derivative of sqrt(x) at x=25.

Answer 15

yes @hsjunior

Answer 16

@Xavier Absolutely

Answer 17

So if you were to encounter this problem, what would you do first?

Answer 18

I'd rationalize it first, multiplying and dividing by the conjugate

Answer 19

so how is it \[\frac{ 1 }{ 10 }\]

Answer 20

You have
\[\lim_{h \to 0}\frac{25+h-25}{h(\sqrt{25+h}+5)}\]
Numerator pops out as h
\[\lim_{h \to 0}\frac{h}{h(\sqrt{25+h}+5)}\]
\[\lim_{h \to 0}\frac{\cancel h}{\cancel h(\sqrt{25+h}+5)}\]
We get
\[\lim_{h\to 0}\frac{1}{\sqrt{25+h}+5}\]
put h=0
\[\frac{1}{\sqrt{25+0}+5}=\frac 1 {5+5}\]

Answer 21

I understand it now. Thanks guys (:

Answer 22

Welcome :)

Answer 23

Do you know all subjects? Is it okay if I ask you a IB Chemistry problem?

Answer 24

I don't know Chemistry much. You should post the question in Chemistry Subject
http://openstudy.com/study#/groups/Chemistry

Answer 25

okay thanks