anonymous
  • anonymous
use L'Hopital's rule to calculate the following derivatives ; lim x->1 x^1/1-x lim x-> x^2-e^x/x
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
the answer for first is 0 . just need the steps.
anonymous
  • anonymous
you mean the following limits?
anonymous
  • anonymous
is the first one \[\lim_{x \rightarrow 1}{x \over 1-x}?\]

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anonymous
  • anonymous
nope
anonymous
  • anonymous
the its x^ 1/1-x
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anonymous
  • anonymous
plz see the attachment its question number 2 -> 5th and 6th one
anonymous
  • anonymous
any answers / suggestions ?
nowhereman
  • nowhereman
you should really learn how to use parentheses! x^1/1-x 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/(1-x)) ?
anonymous
  • anonymous
yup. that's right sorry for the confusion
anonymous
  • anonymous
x^(1/(1-x)) looks good
nowhereman
  • nowhereman
So you should apply the definition of a^b
anonymous
  • anonymous
ok.. 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
  • anonymous
well in l'hopital's rule u have to take drivatives
anonymous
  • anonymous
Consider the function: \[y=x ^{{1 \over 1-x}}\], 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: \[\ln y={1 \over x-1}\ln x={\ln x \over x-1}\] (0/0 type).. now we can apply the rule find the limit for (lny) for now
nowhereman
  • nowhereman
Aha, 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
  • anonymous
hmm..
anonymous
  • anonymous
sorry for the mistake.. it's 1/(1-x), so: \[\lim_{x \rightarrow 1}\ln y=\lim_{x \rightarrow 1}{\ln x \over 1-x}=\lim_{x \rightarrow 1}{1/x \over -1}=-1\] so this\[\lim_{x \rightarrow 1}x ^{{1 \over 1-x}}=e ^{\lim_{x \rightarrow 1 \ln y}}=e ^{-1}={1 \over e}\] the limit of ln y, which implies that
anonymous
  • anonymous
there is a typo!! but the idea is still there.. does that make any sense to you?
anonymous
  • anonymous
just 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
  • anonymous
i mean the equation is in 0 form right
anonymous
  • anonymous
what do you mean by 0 form?
anonymous
  • anonymous
why r u using ln in this
anonymous
  • anonymous
It's the way to finding the limit using l'hopital's rule.. this method is well known.
anonymous
  • anonymous
I am sure your instructor will use the same method to find the limit :P
nowhereman
  • nowhereman
The original term is not in a form where you can apply l'hospitals rule. So you write it as \[e^{\frac{\ln x}{1-x}}\] 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: \[\lim_{x → 1} x^{\frac 1 {1-x}} = \lim_{x→1} e^{\frac{\ln x}{1-x}} = e^{\lim_{x→1} \frac{\ln x}{1-x}} = e^{\lim_{x→1} \frac{\frac{1}{x}}{-1}} = e^{-1} = \frac{1}{e}\]
anonymous
  • anonymous
^^ COOL :)
anonymous
  • anonymous
what about second question
nowhereman
  • nowhereman
What is the value that x goes to? And do you mean x^2 - ( (e^x) / x ) ?
anonymous
  • anonymous
x-> infinity , and yes ur right. it was my writing mistake. as before thanks
anonymous
  • anonymous
x^2 - ( (e^x) / x ) is right!
nowhereman
  • nowhereman
Then 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: \[e^x \geq \underbrace{x^3-Cx}_{\text{polynomial}} ⇔ x^2 - \frac{e^x}{x} \leq C\]
anonymous
  • anonymous
ok

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