Question

If someone could explain the answer to 1J-2 on the first assignment (http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/problem-set-1/MIT18_01SC_pset1prb.pdf) this would be very helpful.

Answer 1

It's impressive that Jerison arrived at this limit in this way, i'm almost in awe!
(Also, i'm not sure whats wrong with this software atm, but it's not displaying fractions very well, the fractions are overwriting the equals signs, so ill put statements on different lines)
Ill explain to you why each statement in the answer is equal to the next one.
We are trying to find the following limit, that is the limit as x approaches pi over 2.
\[\lim_{x \rightarrow \pi/2} \frac{\cos{x}}{x-\pi/2}\]
Now you can observe that
\[\cos{\pi/2} = 0\]
and, that statement is CONSTANT, in other words it's another way of writing zero. So subtracting that from the initial limit doesnt change the value of that limit and we get
\[\lim_{x \rightarrow \pi/2 }\frac{\cos x - \cos{\pi/2}}{x-\pi/2} \]
Now if you look at that limit, and you compare it with the DEFINITION of the derivative and for cosx you will see that they are equivalent, ill give it here
\[\frac{d}{dx} f(x) = \lim_{x \rightarrow \Delta x}\frac{f(x+\Delta x)-f(x)}{\Delta x}\]
\[\frac{d}{dx}\cos(x) = \lim_{\Delta x \rightarrow 0} \frac{\cos(x+\Delta x)-\cos(x)}{\Delta x}\]
Now you can see that the limit now equals the derivative of cosx (when x EQUALS pi divided by 2)
\[\lim_{x \rightarrow \pi/2 }\frac{\cos x - \cos{\pi/2}}{x-\pi/2} \]
Because in the above ( and this is important ) as X APPROACHES pi divided by 2 you can see X MINUS pi / 2 is APPROACHING zero! the same as the limit delta x is APPROACHING zero in the cosx derivative.
\[= \frac{d}{dx} cos(x)|_{x=\pi/2}\]
now because these two are equal you can just take the derivative of cos(x) at pi divided by 2
which as we know is
\[-sin(\pi/2) = -1 \]

Answer 2

Correction above ( you probably caught it ) the derivative definition should be delta x approaching 0 not x approaching delta x

Answer 3

Just to make it crystal clear ( not sure I did a great job above ) ill break it all down and number the relevant parts, ie the (1),(2a),(2b) are not mathematical statements, they just denote the equations and make it easier to refer to the equations.
\[(1):\lim_{x \rightarrow \frac{\pi}{2} } \frac{cos(x)-cos(\frac{\pi}{2})}{x-\frac{\pi}{2}} = (2a):\frac{d}{dx}cos(x)|_{x=\pi/2} = (2b):\lim_{\Delta x \rightarrow 0}\frac{cos(\frac{\pi}{2} + \Delta x)-cos(\frac{\pi}{2})}{\Delta x}\]
In the answer Jerison gives (above) this all relies on the fact that we are using a set value of x in (2b).It must be very clear that the limit in (2b) is actually the DERIVATIVE of cos(x) at
\[x=\frac{\pi}{2}\]
So to show exactly how these two are equal, we will break it all down.
First, in both limits (1) and (2b) both the denominators are approaching zero, that is
in (1) and (2b)
\[(1):x-\frac{\pi}{2}\rightarrow 0, (2b): \Delta x \rightarrow 0\]
shows the equality of the denominators.
Now lets equate the numerators in both limits, first in
\[(1):\lim_{x \rightarrow \frac{\pi}{2}}-cos(\frac{\pi}{2})=0\]
and in (2b) we have
\[(2b): \lim_{\Delta x \rightarrow 0} -cos(\frac{\pi}{2})= 0\]
Again both are equal, now for the last part of the numerator,
in (1)
\[\lim_{x \rightarrow \frac{\pi}{2} }cos(x) \rightarrow cos(\frac{\pi}{2})\]
and in (2b) its approaching the same
\[\lim_{\Delta x \rightarrow 0} cos(\frac{\pi}{2}+\Delta x) \rightarrow cos(\frac{\pi}{2})\]
This shows the reasoning behind Jerisons statement that
\[(1):\lim_{x \rightarrow \frac{\pi}{2} } \frac{cos(x)-cos(\frac{\pi}{2})}{x-\frac{\pi}{2}} = (2a):\frac{d}{dx}cos(x)|_{x=\pi/2}\]

Answer 4

Thanks alot for the detailed answer!
I'm beginning to understand the thought process behind this derivation thanks to your detailed explanation :)

Answer 5

Anytime