A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
use L'Hopital's rule to calculate the following derivatives ;
lim x>1 x^1/1x
lim x> x^2e^x/x
anonymous
 5 years ago
use L'Hopital's rule to calculate the following derivatives ; lim x>1 x^1/1x lim x> x^2e^x/x

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer for first is 0 . just need the steps.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean the following limits?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the first one \[\lim_{x \rightarrow 1}{x \over 1x}?\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0plz see the attachment its question number 2 > 5th and 6th one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0any answers / suggestions ?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0you should really learn how to use parentheses! x^1/1x means ((x^1)/1)x because ^ has higher priority than / (and *) which in turn has higher priority than  (and +) ! So do you mean x^(1/(1x)) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yup. that's right sorry for the confusion

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^(1/(1x)) looks good

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0So you should apply the definition of a^b

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok.. first the question is clearly asking about the limits, not the derivatives ( I think it's just a typo mistake).. I will start with the first one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well in l'hopital's rule u have to take drivatives

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Consider the function: \[y=x ^{{1 \over 1x}}\], which is the function we are going to find its limit as it goes to 1.. we can't use l'hopital's rule since it's not in the form of (0/0) or (infinity/infinity).. so we are going to take the ln for for both side to get: \[\ln y={1 \over x1}\ln x={\ln x \over x1}\] (0/0 type).. now we can apply the rule find the limit for (lny) for now

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Aha, applying ln, so where does that come from? Magic, magic. Just use the definition \[a^b = e^{b\ln a}\] and you will get exactly this result.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry for the mistake.. it's 1/(1x), so: \[\lim_{x \rightarrow 1}\ln y=\lim_{x \rightarrow 1}{\ln x \over 1x}=\lim_{x \rightarrow 1}{1/x \over 1}=1\] so this\[\lim_{x \rightarrow 1}x ^{{1 \over 1x}}=e ^{\lim_{x \rightarrow 1 \ln y}}=e ^{1}={1 \over e}\] the limit of ln y, which implies that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there is a typo!! but the idea is still there.. does that make any sense to you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just because we took ln for the limit of ln y.. to get the limit of y, we just took e to the power of that first limit.. as simple as that!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mean the equation is in 0 form right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you mean by 0 form?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why r u using ln in this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's the way to finding the limit using l'hopital's rule.. this method is well known.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am sure your instructor will use the same method to find the limit :P

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0The original term is not in a form where you can apply l'hospitals rule. So you write it as \[e^{\frac{\ln x}{1x}}\] where the exponent is in the form "0/0". Then you find the limit of the exponent with l'hopsital as 1 and thus have 1/e as the end result because the exponential function is continuous: \[\lim_{x → 1} x^{\frac 1 {1x}} = \lim_{x→1} e^{\frac{\ln x}{1x}} = e^{\lim_{x→1} \frac{\ln x}{1x}} = e^{\lim_{x→1} \frac{\frac{1}{x}}{1}} = e^{1} = \frac{1}{e}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what about second question

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0What is the value that x goes to? And do you mean x^2  ( (e^x) / x ) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x> infinity , and yes ur right. it was my writing mistake. as before thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2  ( (e^x) / x ) is right!

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Then it diverges to ∞, because e^x grows faster than any polynomial. So for any C ϵ ℝ you can find a X so that for all x > X you have: \[e^x \geq \underbrace{x^3Cx}_{\text{polynomial}} ⇔ x^2  \frac{e^x}{x} \leq C\]
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.