Question

f(2 + h) − f(2)/h where f(x) = (x+1)^1/2

Answer 1

limits as h approaches 0 ?

Answer 2

yes

Answer 3

i just cant figure out what to do with the 2?

Answer 4

for get the 2 and just derive f(x)

Answer 5

right.. this is about differentiating \[\sqrt{x+1}\] at x=2.

so use L hospitals rule.. on the limit below

\[(\sqrt{3+h} - \sqrt{3} ) \div h\] as h approaches zero.

Answer 6

\[(x+1)^{1/2}=\sqrt{x+1}\]
\[\lim_{h->0}\frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}\text{ ; use the conjugate}\]

Answer 7

this 'll be \[1\div( 2\sqrt{3+h})\] with h=0. ie.. 1/2sqrt3

Answer 8

yea i got that thanks

Answer 9

\[\lim_{h->0}\frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}*\frac{\sqrt{x+h+1}+\sqrt{x+1}}{\sqrt{x+h+1}+\sqrt{x+1}}\]
\[\lim_{h->0}\frac{x+h+1-{x-1}}{h(\sqrt{x+h+1}+\sqrt{x+1})}\]
\[\lim_{h->0}\frac{h}{h(\sqrt{x+h+1}+\sqrt{x+1})}\]
\[\lim_{h->0}\frac{1}{(\sqrt{x+h+1}+\sqrt{x+1})}\]
now when h=0 we get
\[\frac{1}{\sqrt{x+1}+\sqrt{x+1}}\]
\[\frac{1}{2\sqrt{x+1}}\]
and solve for x=2

Answer 10

isn't just differentiating the numerator & denominator easier?

Answer 11

its not about easier at the beginning of calculus1 ; its about getting a stong foundation that prepares you to trust the "shortcuts" :)

Answer 12

sqrt(x) is an elementary function; which leads me to believe that this isnt really about apply Lhopitals rule to solve ;)

Answer 13

touche.