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  1. Umrao

    Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}##

    Cross checked with wolfram alpha, it says that h(x) is discontinuous at x=0.
  2. Umrao

    Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}##

    I didn't put much thought to what was implied by differentiation (sorry I'll try to figure it out.) My book says: "A function f(x) is said to be continuous at x = a; where a is an element of domain of f(x). If left hand limit at a = right hand limit at a = f(a) i.e. LHL=RHL=Value of function at...
  3. Umrao

    Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}##

    I think h is not continous at 0. limit h(0) does exist but limit has to be equal to h(0) which is undefined. h(0) = infinity/infinty. You had h(0) = ln2 by taking the limit (using a standard formula.)
  4. Umrao

    Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}##

    Finally! Thanks a ton! Yipee! got the answer. By the way, I was curious from your first post (on this thread) - what does - "mfb" stand for? Your initials? Praise be unto He. Helium!!
  5. Umrao

    Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}##

    When I was posting the last thread, it was 5am here, and so I was feeling very sleepy, so I thought I am not getting it right now, I should try later. But even now that I have tried when I am wide awake I can't find the two products you mentioned. As in my last thread The two factors that I...
  6. Umrao

    Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}##

    I am sorry but I could not pick up the clue. From your description I got. $$1/4 \lim_{y/2\to 0}[(\frac {e^{y/2}-y/2-1}{y/2})(\frac {e^{y/2}-y/2+1}{y/2})-1]$$ Am I going right?
  7. Umrao

    Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}##

    Homework Statement Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}## without using L'Hospital's rule or expansion of the series. Answer is given to be = ##\frac{(\log_e (2))^2}{2}## Homework Equations Squeeze play theorem/ Sandwich theorem, some algebraic manipulations and standard...
  8. Umrao

    Integration using Leibnitz theorem

    Put the damn value of c to get that. f(0) = 0
  9. Umrao

    Integration using Leibnitz theorem

    Why don't you tell me where
  10. Umrao

    Integration using Leibnitz theorem

    Then why did you bother me with all the latex. I wasted 1.5 hours
  11. Umrao

    Integration using Leibnitz theorem

    Finally, deciphered. Yet another mistake in calculation and hopefully this will be my last thread to this post. Mistake -1. ##f'(cos\phi)=\frac{2}{\sin\phi}\cdot(\arctan\cot\phi-\arctan\tan\phi)=\frac{\pi}{\sin\phi}##. Since R.H.S. is correct so you made a typo on computer here (started...
  12. Umrao

    Integration using Leibnitz theorem

    I hope you do know $$\tan^{-1} (cot \phi) = \tan^{-1}(tan (\frac{\pi}{2}-\phi))= \frac{\pi}{2}-\phi$$
  13. Umrao

    Integration using Leibnitz theorem

    Why did you put that proof here? Are you even reading what I am writing? Look I don't understand what are you trying to say. If your confusion is cleared then good, if it is still there, then I am going to ask you to elaborate it, clearly. And by the way why did you put that proof here?
  14. Umrao

    Integration using Leibnitz theorem

    I see the problem there. Make a right angled triangle, take an angle as phi. Since sum of angles of triangle is pi hence the other angle has to be pi/2 - phi. Find tan of phi is perpendicular upon base [Let's say = AB/BC, ABC is triangle and B is right angle and C is phi]. cot of (pi/2 -phi) is...
  15. Umrao

    Integration using Leibnitz theorem

    Don't put $$\arctan (tan (\frac{\pi}{2} - \phi))$$ as $$(\arctan\cot\phi)$$ I wrote $$\arctan (tan (\frac{\pi}{2} - \phi))$$ deliberately, cause I have some devious plans for it. Apply some manipulation (like taking out ph...):wink:
  16. Umrao

    Integration using Leibnitz theorem

    You made a mistake. $$f'(b)=\frac{2}{\sqrt{1-b^2}}\big[\arctan{\frac{\sqrt{1+b}}{\sqrt{1-b}}}+\arctan{\frac{\sqrt{1-b}}{\sqrt{1+b}}}\big] $$ After this is your mistake. $$ \rightarrow \...
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