Question

How do you solve final exam question 3b? lim as x ->3^.5 of (arctan(x) - pi/3) / (x - 3^.5)?

Answer 1

There are two ways, one is really quick the other not so quick.
The not so quick way is to compute values closer and closer to the value x approaches, in this case its roughly \[\sqrt{3} = 1.732050808\]
Lets take three values that get closer and closer to the square root of 3 from the positive side.
Say 1.78, 1.73 and 1.73206 and plug them into
\[\frac{arctan(x)-\frac{\pi}{3}}{x-\sqrt{3}}\]
we get
0.244903669
0.249897284
0.249999005
if you keep doing this (from both sides) you will notice that you can never quite get to the value 0.25 thus creating this limit:
\[lim_{x \to \sqrt{3}}\frac{arctan(x)-\frac{\pi}{3}}{x-\sqrt{3}}=0.25\]
infact this is how I first found this limit.

The nice and easy way is to use L'Hôpital's rule for limits, and that is, if the limit looks like
\[\frac{0}{0}\] then to calculate the limit take the derivative of the numerator and the derivative of the denominator and plug in the value that the limit is approaching.
In this case the derivative of the numerator is
\[\frac{d}{dx}arctan(x)-\frac{\pi}{3} = \frac{1}{1+x^2}\]
and the derivative of the denominator is
\[\frac{d}{dx}x-\sqrt{3}= 1\]
now we have
\[\frac{1}{\frac{1+x^2}{1}}= \frac{1}{1+x^2}\]
plug
\[\sqrt{3}\] into that equation
and you have
\[\frac{1}{1+\sqrt{3}^2}= \frac{1}{4}\] which you will find is the limit as x approaches the square root of three.

Answer 2

makes sense, thanks