## 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

1. welshfella

2 * e^(pi/2) * cos (pi/2) make sure to set your calculator to radians

2. SolomonZelman

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

3. Astrophysics

What does cos(pi/2) give you?

4. Astrophysics

That should be a give away

5. 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 $$\large f(a)$$.

6. 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....)

7. SolomonZelman

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

8. 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=±∞)

9. anonymous

i got c is that correct? @SolomonZelman @Astrophysics

10. SolomonZelman

$$\cos(90)=?$$

11. 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)

12. anonymous

1?

13. SolomonZelman

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

14. anonymous

0

15. SolomonZelman

yes, this limit is equivalent to zero.

16. anonymous

thank u!!

17. SolomonZelman

(yes, 0, from both sides.)

18. SolomonZelman

Welcome into the calculus world:) xD Good luck with your math!