# Limits Question

• Jan 26th 2010, 10:58 PM
Lord Darkin
Limits Question
If I have this problem:

lim as h -> 0
$\displaystyle \frac{sin( \frac{\pi}{2} + h) - 1}{h}$

And I have these answer choices:

A. $\displaystyle f'(\frac{\pi}{2}), f(x)=sinx$
B. $\displaystyle f'(\frac{\pi}{2}), f(x)=\frac{sinx}{x}$
C. $\displaystyle f'(1), f(x)=sinx$
D. $\displaystyle f'(1), f(x)=sin(x + \frac{\pi}{2})$
E. $\displaystyle f'(0), f(x)=sinx$

What does this really mean? like, I don't understand what f'(0) would imply, for instance.
• Jan 27th 2010, 02:00 AM
tonio
Quote:

Originally Posted by Lord Darkin
If I have this problem:

lim as h -> 0
$\displaystyle \frac{sin( \frac{\pi}{2} + h) - 1}{h}$

And I have these answer choices:

A. $\displaystyle f'(\frac{\pi}{2}), f(x)=sinx$
B. $\displaystyle f'(\frac{\pi}{2}), f(x)=\frac{sinx}{x}$
C. $\displaystyle f'(1), f(x)=sinx$
D. $\displaystyle f'(1), f(x)=sin(x + \frac{\pi}{2})$
E. $\displaystyle f'(0), f(x)=sinx$

What does this really mean? like, I don't understand what f'(0) would imply, for instance.

If a function f(x) is defined on a point $\displaystyle x_0$ and in some neighborhood of this point , the definition of the derivative of f in this point is:

$\displaystyle f'(x_0):=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$ , in case this limits exists and is finite.

Can you see now what option to choose?

Tonio
• Jan 27th 2010, 02:39 AM
Lord Darkin
So ... $\displaystyle f(x_{0}) = 1$

Would it be choice a? Since if I put in pi/2, then that means that f(x) would equal 1 since sin(90)=1.
• Jan 27th 2010, 03:53 AM
HallsofIvy
Yes, that is correct.

(Although I would prefer "$\displaystyle sin(\pi/2)= 1$" to "sin(90)= 1"! You need to start thinking in radians, not degrees.)
• Jan 27th 2010, 05:59 AM
Lord Darkin
Yeah, I would ahve thought that. Thanks. :)