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September 8th, 2017, 09:38 PM   #1
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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 11:58 PM.
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September 8th, 2017, 10:07 PM   #2
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switch from degrees to radians on your calculator
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September 8th, 2017, 10:11 PM   #3
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Thanks a lot. that was the problem and yes the numbers match now. Thanks a lot.
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September 8th, 2017, 10:42 PM   #4
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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/2-arcsine?
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September 8th, 2017, 11:16 PM   #5
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Quote:
Originally Posted by Leonardox View Post
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/2-arcsine?
$\dfrac \pi 2$ is used because it is the limit of the arctan function as $x^2 \to \infty$

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.
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September 8th, 2017, 11:45 PM   #6
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Got it thanks
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