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

- 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

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- anonymous

the answer for first is 0 . just need the steps.

- anonymous

you mean the following limits?

- anonymous

is the first one
\[\lim_{x \rightarrow 1}{x \over 1-x}?\]

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## More answers

- anonymous

nope

- anonymous

the its x^ 1/1-x

##### 1 Attachment

- anonymous

plz see the attachment its question number 2 -> 5th and 6th one

- anonymous

any answers / suggestions
?

- 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

yup. that's right
sorry for the confusion

- anonymous

x^(1/(1-x)) looks good

- nowhereman

So you should apply the definition of a^b

- 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

well in l'hopital's rule u have to take drivatives

- 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

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

hmm..

- 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

there is a typo!! but the idea is still there.. does that make any sense to you?

- 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

i mean the equation is in 0 form right

- anonymous

what do you mean by 0 form?

- anonymous

why r u using ln in this

- anonymous

It's the way to finding the limit using l'hopital's rule.. this method is well known.

- anonymous

I am sure your instructor will use the same method to find the limit :P

- 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

^^
COOL :)

- anonymous

what about second question

- nowhereman

What is the value that x goes to? And do you mean x^2 - ( (e^x) / x ) ?

- anonymous

x-> infinity , and yes ur right. it was my writing mistake. as before thanks

- anonymous

x^2 - ( (e^x) / x ) is right!

- 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

ok

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