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Mathematics 9 Online
OpenStudy (anonymous):

Find the limit of the function by using direct substitution.

OpenStudy (anonymous):

OpenStudy (anonymous):

The choices are: a. 1 b. pi/2 c. 0 d. 5e^pi/2

OpenStudy (rock_mit182):

@iPwnBunnies @ganeshie8

OpenStudy (ipwnbunnies):

Direct substitution means we can find the limit by plugging in the value of the limit into x. If it's a real number, that will be the limit as x approaches pi/2 of the function.

OpenStudy (anonymous):

I'm confused about the e

OpenStudy (rock_mit182):

so am i

OpenStudy (anonymous):

@zimmah

OpenStudy (campbell_st):

well if you use direct substitute you get \[3e^{\frac{\pi}{2}} \times \cos(\frac{\pi}{2})\] so it should be easy to solve from here

OpenStudy (rock_mit182):

i mean i didn't undestand the graph of 3e to the pith 0ver2

OpenStudy (anonymous):

Yeah I don't know what to do with 3e^pi/2

OpenStudy (campbell_st):

you don't need to know about \[3e^{\frac{\pi}{2}}\] because \[\cos(\frac{\pi}{2}) = 0\] so in evaluating the limit you have \[3e^{\frac{\pi}{2}} \times \cos(\frac{\pi}{2}) = 3e^{\frac{\pi}{2}} \times 0 \]

OpenStudy (anonymous):

So that means it is 0

OpenStudy (anonymous):

correct, i just arrived home so sorry i couldn't reply earlier but campbell is correct.

OpenStudy (anonymous):

the e by the way is similar to pi, except pi is approximately 3.1415926... and e is approximately 2.71828...

OpenStudy (rock_mit182):

if only we have the e part how the limit look like ?

OpenStudy (campbell_st):

well just graph the curve \[y = 3e^xcos(x)\]

OpenStudy (rock_mit182):

i guess there is an exact value

OpenStudy (rock_mit182):

\[\lim_{x \rightarrow \pi/2} = 3e ^{x}\]

OpenStudy (campbell_st):

here is your graph

OpenStudy (campbell_st):

if you want to graph it yourself use https://www.desmos.com/calculator

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