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...
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.)
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!!
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...
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?
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...
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...
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?
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...
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:
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 \...