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
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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!

Looking for something else?

Not the answer you are looking for? Search for more explanations.