Question

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

lim x-> x^2-e^x/x

Answer 1

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

Answer 2

you mean the following limits?

Answer 3

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

Answer 4

nope

Answer 5

the its x^ 1/1-x

Answer 6

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

Answer 7

any answers / suggestions
?

Answer 8

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)) ?

Answer 9

yup. that's right
sorry for the confusion

Answer 10

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

Answer 11

So you should apply the definition of a^b

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

hmm..

Answer 17

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

Answer 18

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

Answer 19

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!!

Answer 20

i mean the equation is in 0 form right

Answer 21

what do you mean by 0 form?

Answer 22

why r u using ln in this

Answer 23

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

Answer 24

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

Answer 25

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: \[\lim_{x → 1} x^{\frac 1 {1-x}} = \lim_{x→1} e^{\frac{\ln x}{1-x}} = e^{\lim_{x→1} \frac{\ln x}{1-x}} = e^{\lim_{x→1} \frac{\frac{1}{x}}{-1}} = e^{-1} = \frac{1}{e}\]

Answer 26

^^
COOL :)

Answer 27

what about second question

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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: \[e^x \geq \underbrace{x^3-Cx}_{\text{polynomial}} ⇔ x^2 - \frac{e^x}{x} \leq C\]

Answer 32

ok