September 8th, 2017, 08:38 PM  #1 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6  Epsilon Delta for infinity
Hello Guys, I need someone to explain me the answer. Especially, how did we get 1.526 using arctan 22.36? I used the arctan function of the calculator and typed arctan((22.36) and I did not get the same numbers. Please explain to me how 1.526 and 0.0447 are obtained. I really appreciate. Question: Given that $\displaystyle \lim_{x\to\pi/2} \tan^2\!x = \infty$, illustrate the definition by finding values of δ that correspond to the following. (Round your answer down to four decimal places.) M=500 M=1000 Answer: x > arctan √500 ~ arctan 22.37 ~ 1.526 δ =  pi/2  arctan √500 = 0.0447 example take δ = 0.04 and x = 1.5307 (tan 1.5307)^2 ~ 621.334 > 500 if M=1000 δ =  pi/2  arctan √1000 ~ 0.032 take x = 1.54 (tan 1.54)^2 ~ 1053.7 > 1000 Last edited by skipjack; September 8th, 2017 at 10:58 PM. 
September 8th, 2017, 09:07 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,943 Thanks: 1009 
switch from degrees to radians on your calculator

September 8th, 2017, 09:11 PM  #3 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 
Thanks a lot. that was the problem and yes the numbers match now. Thanks a lot.

September 8th, 2017, 09:42 PM  #4 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 
in the line "(pi/2)arctan(whatever the value)" pi/2  arctan rule? or pi/2 is because of tanx^2. if my function was sin cos or anything would I still do pi/2arcsine? 
September 8th, 2017, 10:16 PM  #5  
Senior Member Joined: Sep 2015 From: USA Posts: 1,943 Thanks: 1009  Quote:
Sine and Cosine in particular have no such limits as they are periodic. For other functions, you would use their limits as their arguments tend to infinity if such limits exist.  
September 8th, 2017, 10:45 PM  #6 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 
Got it thanks


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