## anonymous 5 years ago use L'Hopital's rule to calculate the following derivatives ; lim x->1 x^1/1-x lim x-> x^2-e^x/x

1. anonymous

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

2. anonymous

you mean the following limits?

3. anonymous

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

4. anonymous

nope

5. anonymous

the its x^ 1/1-x

6. anonymous

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

7. anonymous

8. 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)) ?

9. anonymous

yup. that's right sorry for the confusion

10. anonymous

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

11. nowhereman

So you should apply the definition of a^b

12. 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

13. anonymous

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

14. 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

15. 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.

16. anonymous

hmm..

17. 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

18. anonymous

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

19. 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!!

20. anonymous

i mean the equation is in 0 form right

21. anonymous

what do you mean by 0 form?

22. anonymous

why r u using ln in this

23. anonymous

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

24. anonymous

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

25. 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: $\lim_{x → 1} x^{\frac 1 {1-x}} = \lim_{x→1} e^{\frac{\ln x}{1-x}} = e^{\lim_{x→1} \frac{\ln x}{1-x}} = e^{\lim_{x→1} \frac{\frac{1}{x}}{-1}} = e^{-1} = \frac{1}{e}$

26. anonymous

^^ COOL :)

27. anonymous

28. nowhereman

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

29. anonymous

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

30. anonymous

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

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

32. anonymous

ok