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anonymous

  • one year ago

Find the limit of the function by using direct substitution. lim-> pi/2 (2e^x cosx) 0 1 2e^(pi/2) pi/2

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  1. welshfella
    • one year ago
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    2 * e^(pi/2) * cos (pi/2) make sure to set your calculator to radians

  2. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \lim_{x\rightarrow \pi/2}~2e^x\cos(x) =2e^{\pi/2}\cos(\pi/2) }\)

  3. Astrophysics
    • one year ago
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    What does cos(pi/2) give you?

  4. Astrophysics
    • one year ago
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    That should be a give away

  5. SolomonZelman
    • one year ago
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    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 \(\large f(a)\).

  6. SolomonZelman
    • one year ago
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    (most limits as you have previously seen or going to see, you can't apply direct substitution tho', whether that is fortunate or not....)

  7. SolomonZelman
    • one year ago
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    (( We can do a taylor approximation here for \(\pi/2\) and for e, if you would like ))

  8. SolomonZelman
    • one year ago
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    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=±∞)

  9. anonymous
    • one year ago
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    i got c is that correct? @SolomonZelman @Astrophysics

  10. SolomonZelman
    • one year ago
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    \(\cos(90)=?\)

  11. SolomonZelman
    • one year ago
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    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)

  12. anonymous
    • one year ago
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    1?

  13. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \lim_{x\rightarrow \pi/2}~2e^x\cos(x) =2e^{\pi/2}\cdot \cos(\pi/2) =2e^{\pi/2}\cdot 0=?}\)

  14. anonymous
    • one year ago
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    0

  15. SolomonZelman
    • one year ago
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    yes, this limit is equivalent to zero.

  16. anonymous
    • one year ago
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    thank u!!

  17. SolomonZelman
    • one year ago
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    (yes, 0, from both sides.)

  18. SolomonZelman
    • one year ago
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    Welcome into the calculus world:) xD Good luck with your math!

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