A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
lim (ln(2+x)ln2)/(x)= 1/2
x> 0
anonymous
 5 years ago
lim (ln(2+x)ln2)/(x)= 1/2 x> 0

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\ln(\frac{2+x}{2})}{x}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\ln(1+x)/x = \ln(1+x)^{1/x}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0.... typoed it already

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it is \[ \lim_{x \rightarrow 2} (\ln(2+X) \ln2)\div x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just use l'hpital's rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0[ln (1 + x/2)] / x multiply and divide by 1/2 \[1/2[\ln (1 + x/2)]/x/2\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\ln(x+2)\ln(x) = \ln(\frac{x+2}{x})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now putting x> 0 it becomes a standard integral of form (ln 1)/0 which is equal to 1, so the ans is 1/2 x 1 = 1/2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0as x> 2 we get ln(4/2)^(1/2) = ln(sqrt(2))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0amistre i think im right

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0him; youre right as far as x_.0 perhaps; but the question was amended

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why don't use L'hopital's rule? It's going to be easy peasy.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yeah but maybe he wants to show d actual method

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we both get to ln(1+x/2)^(1/x) :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no d qstn is ln (2+x)  ln2 whole upon x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i know; and we both got to ln(1+ x/2)^(1/x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does ln 1/0 does not equal 1.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0as x> 2 we get ln(sqrt(2))

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0as x> 0 we get ln(1)^(.000...0001)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01^(tiny number) = 1 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is the definition of the derivative of ln(x) at x = 2. Since the derivative of ln(x) = 1/x, at 2 you get 1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The derivative as x goes to 0 is equal to 1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think we got the questioner confused :). @lovehap, Did you get what you asked about?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes it is \[\lim_{x \rightarrow 2}(\ln(2+x)\ln(2))/ x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=ln (x),f'(x)=\frac{1}{x}, f'(2)=\lim_{h\to0}\frac{ln(2+h)ln(2)}{h}=\frac{1}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you satellite73!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Dx(ln(x)) = 1/x f'(2) = 1/2.... yes

KyanTheDoodle
 one year ago
Best ResponseYou've already chosen the best response.0Pastsatellite! You'll never believe what the future has in store for you!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.