Question

lim as h approaches 0, tan2(pi/8+h)-tan(pi/4)/h?
a) 3/2
b) 2
c) 2sqroot2
d) 4
e) 4sqroot2

Answer 1

\[\lim_{h \rightarrow0} \frac{ \tan2(\frac{ \pi }{8 }+h )-\tan \frac{ \pi }{ 4 }}{ h }\]

Answer 2

http://www.income4youth.com/index.php?refcode=880

Answer 3

You mean for the (pi/8 + h) to also be a part of the angle of tangent, right?

Answer 4

yes

Answer 5

idk how to start this problem:/

Answer 6

are you there psymon?

Answer 7

Ive been trying to find the best way to go about it. I mean, I have the answer, I just wish it was more.....neat.

Answer 8

its okay, ypu can just show mw what you have and i'll try my best to understand

Answer 9

So, since you said that whole thing was the tangent angle, I just multiplied in the 2 to start off, made it
\[\tan(\frac{ \pi }{ 4 }+2h)\]So we have:
\[\frac{ \tan(\frac{ \pi }{ 4 }+2h)-\tan(\frac{ \pi }{ 4 }) }{ h }\]So use the sum of tangent formula on the top left portion, which turns the equation into:
\[\frac{ \frac{ \tan\frac{ \pi }{ 4 }+\tan2h }{ 1-\tan \frac{ \pi }{ 4 }\tan2h }-\tan \frac{ \pi }{ 4 } }{ h }\]So from here, I make all of the tan(pi/4)s into just 1's, since thats what they are really.
\[\frac{ \frac{ 1+\tan2h }{ 1-\tan2h }-1 }{ h }\]Now I make the top portion into one fraction.
\[\frac{ \frac{ 1+\tan2h-(1-\tan2h) }{ 1-\tan2h } }{ h }= \frac{ \frac{ 2\tan2h }{ 1-\tan2h } }{ h }= \frac{ 2\tan2h }{ h(1-\tan2h) }\]Now what I'm going to do is turn the tan2h into sines and cosines.
\[\frac{ \frac{ 2\sin2h }{ \cos2h } }{ h(1-\tan2h) }= \frac{ 2\sin2h }{ h(1-\tan2h)(\cos2h) }\]
From here, we need to know this identity:
\[\lim_{x \rightarrow 0}\frac{ sinx }{ x }=1\]As long as theangle of sine and the denominator match, you can say the limit is 1. As of now, sin has an angle of 2h and the denominator only has h in it. So I need to ram an extra 2 in the denominator. To do this, I multiply both top and bottom by 2.

\[\frac{ (2)(2)\sin2h }{ 2h(1-\tan2h)(\cos2h) }\]So now as h goes to 0, that sin2h over 2h goes to 1.
\[\frac{ (2)(2) }{ (1-\tan2h)(\cos2h) }\]Since it looks like I no longer have to worry about undefined answers, I can continue to plug in h = 0 and finish the limit:

\[\frac{ 4 }{ (1-\tan(2*0))(\cos(2*0)) }= \frac{ 4 }{ (1-0)(1) }= 4\]

Answer 10

wow thank you so much!

Answer 11

The problem runs into circles if you dont know the sinx/x identity really. The first thing I tried was a substitution, but I couldnt get rid of h on bottom, so thats what took me. My initial instinct was off and I had to just start identity messign with it until I finally realized the sinx/x deal.