use L'Hopital's rule to calculate the following derivatives ; lim x->1 x^1/1-x lim x-> x^2-e^x/x

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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
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the answer for first is 0 . just need the steps.
you mean the following limits?
is the first one \[\lim_{x \rightarrow 1}{x \over 1-x}?\]

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Other answers:

nope
the its x^ 1/1-x
1 Attachment
plz see the attachment its question number 2 -> 5th and 6th one
any answers / suggestions ?
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)) ?
yup. that's right sorry for the confusion
x^(1/(1-x)) looks good
So you should apply the definition of a^b
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
well in l'hopital's rule u have to take drivatives
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
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.
hmm..
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
there is a typo!! but the idea is still there.. does that make any sense to you?
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!!
i mean the equation is in 0 form right
what do you mean by 0 form?
why r u using ln in this
It's the way to finding the limit using l'hopital's rule.. this method is well known.
I am sure your instructor will use the same method to find the limit :P
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: \[\lim_{x → 1} x^{\frac 1 {1-x}} = \lim_{x→1} e^{\frac{\ln x}{1-x}} = e^{\lim_{x→1} \frac{\ln x}{1-x}} = e^{\lim_{x→1} \frac{\frac{1}{x}}{-1}} = e^{-1} = \frac{1}{e}\]
^^ COOL :)
what about second question
What is the value that x goes to? And do you mean x^2 - ( (e^x) / x ) ?
x-> infinity , and yes ur right. it was my writing mistake. as before thanks
x^2 - ( (e^x) / x ) is right!
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: \[e^x \geq \underbrace{x^3-Cx}_{\text{polynomial}} ⇔ x^2 - \frac{e^x}{x} \leq C\]
ok

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