anonymous
  • anonymous
Find the limit of the function by using direct substitution. lim-> pi/2 (2e^x cosx) 0 1 2e^(pi/2) pi/2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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welshfella
  • welshfella
2 * e^(pi/2) * cos (pi/2) make sure to set your calculator to radians
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \lim_{x\rightarrow \pi/2}~2e^x\cos(x) =2e^{\pi/2}\cos(\pi/2) }\)
Astrophysics
  • Astrophysics
What does cos(pi/2) give you?

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Astrophysics
  • Astrophysics
That should be a give away
SolomonZelman
  • SolomonZelman
This is what direct substitution means here. When you have a limit \(\large\color{black}{ \displaystyle \lim_{x\rightarrow a}~f(x) }\) then, by direct substitution, your limit is equal to \(\large f(a)\).
SolomonZelman
  • SolomonZelman
(most limits as you have previously seen or going to see, you can't apply direct substitution tho', whether that is fortunate or not....)
SolomonZelman
  • SolomonZelman
(( We can do a taylor approximation here for \(\pi/2\) and for e, if you would like ))
SolomonZelman
  • SolomonZelman
I will tell you this. If you have: \(\large\color{black}{ \displaystyle \lim_{x\rightarrow a}~f(x) }\) then, for any continous function f(x), direct substitution will work. (unless you have a=±∞)
anonymous
  • anonymous
i got c is that correct? @SolomonZelman @Astrophysics
SolomonZelman
  • SolomonZelman
\(\cos(90)=?\)
SolomonZelman
  • SolomonZelman
if you don't remember, then, unit circle, or: \(\color{black}{ \displaystyle \cos(90)=\cos(45+45)=\cos(45)\cos(45)-\sin(45)\sin(45)=0 }\) (knowing that `sin(45)=cos(45)` => = √2/2)
anonymous
  • anonymous
1?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \lim_{x\rightarrow \pi/2}~2e^x\cos(x) =2e^{\pi/2}\cdot \cos(\pi/2) =2e^{\pi/2}\cdot 0=?}\)
anonymous
  • anonymous
0
SolomonZelman
  • SolomonZelman
yes, this limit is equivalent to zero.
anonymous
  • anonymous
thank u!!
SolomonZelman
  • SolomonZelman
(yes, 0, from both sides.)
SolomonZelman
  • SolomonZelman
Welcome into the calculus world:) xD Good luck with your math!

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