Question

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

0
1
2e^(pi/2)
pi/2

Answer 1

2 * e^(pi/2) * cos (pi/2)

make sure to set your calculator to radians

Answer 2

\(\large\color{black}{ \displaystyle \lim_{x\rightarrow \pi/2}~2e^x\cos(x) =2e^{\pi/2}\cos(\pi/2) }\)

Answer 3

What does cos(pi/2) give you?

Answer 4

That should be a give away

Answer 5

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 \(\large f(a)\).

Answer 6

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

Answer 7

(( We can do a taylor approximation here for \(\pi/2\) and for e, if you would like ))

Answer 8

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=±∞)

Answer 9

i got c is that correct? @SolomonZelman @Astrophysics

Answer 10

\(\cos(90)=?\)

Answer 11

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)

Answer 12

1?

Answer 13

\(\large\color{black}{ \displaystyle \lim_{x\rightarrow \pi/2}~2e^x\cos(x) =2e^{\pi/2}\cdot \cos(\pi/2) =2e^{\pi/2}\cdot 0=?}\)

Answer 14

0

Answer 15

yes, this limit is equivalent to zero.

Answer 16

thank u!!

Answer 17

(yes, 0, from both sides.)

Answer 18

Welcome into the calculus world:) xD

Good luck with your math!