Question

Evaluate the following limit. Lim x->π cos2x+5cosx+4 / cosx+1

Answer 1

All of it is divided by cosx + 1, right?

Answer 2

yes

Answer 3

Heres a start for ya then. Change cos2x using the double angle formula identity. Use the identity for it that creates a cos^2, then use that cos^2 to factor the top some. You can force part of the numerator to become quadratic in cosine.

Answer 4

im still confused

Answer 5

Do you know what cos2x identity Im talking about?

Answer 6

no

Answer 7

Yeah, it looks like this problem requires you to know the identity.
\[\cos2x = \cos^{2}x - \sin^{2}x\]

Answer 8

how do i use this in the problem?

Answer 9

You use the cos^2(x) to factor a portion of the top
\[\frac{ -\sin^{2}x + \cos^{2}x +5cosx + 4 }{ cosx+1 }\]If you ignore the sin^2 up there, the cos^2x + 5cosx + 4 can be factored like a quadratic.

Answer 10

im so confused.. like really... I feel so lost..

Answer 11

So do you have any idea how I even came to write it the way I did? Forget solving it, do you understand how I changed it?

Answer 12

i think..

Answer 13

All I did was change cos2x into cos^2(x) - sin^2(x), because it is an identity. Do you know how to factor quadratics? Like, could you factor

x^2 + 5x +6 if you had to?

Answer 14

yes

Answer 15

Okay, awesome. Then you can do that with the top. So the top includes
\[\cos^{2}x + 5cosx +4\]Well, this is the same thing. Just because it has cosine doesnt change the fact that its ax^2 +bx + c. So factor it the same exact way you would as if it were
\[x^{2} + 5x + 4\]

Answer 16

(x+1)(x+4)

Answer 17

Bingo. Well, its just those x's are cosines instead. So now I rewrite the limit like this:
\[\frac{ -\sin^{2}x + (cosx+1)(cosx+4) }{ cosx+1 }\]
Now you can kinda see where this problem might be going : )

Answer 18

ok, so the answer is -sin2x+cosx+4

Answer 19

Well, this is a limit problem, we're not done yet. You can't simplify it in that way, but we're getting there. So now I'm going to take that limit and split it into two fractions. Every term in the numerator can be made into its own fraction. So I can put this:

\[\frac{ -\sin^{2}x }{ cosx+1 }+\frac{ (cosx+1)(cosx+4) }{ (cosx+1) }\]
Can you seehow I did that?

Answer 20

yes so you can't cancel out the like terms?

Answer 21

Well, you cant because there has to be the same like term in every part of the numerator. There was a cosx + 1 in the right term, but the sin^2x portion had no cos to cancel out, so we couldnt do that. But when I split it up into two fractions, that right fraction can cancel out like terms. So that gives me:
\[\frac{ -\sin^{2}x }{ cosx+1 }+cosx+4\]
So this is a start, but we still have that cosx + 1 in the other fraction that will give us problems. We have to find a way to cancel that out, too. I don't suppose you have any idea, huh?

Answer 22

i kinda don't..

Answer 23

my teacher did not explain this well.

Answer 24

Do you remember this identity by chance?

\[\sin^{2}x+\cos^{2}x = 1\]

Answer 25

no

Answer 26

A lot of these things are forcing you to recall knowledge from trig. There is no way to avoid this stuff, unfortunatrly. If you have to do too many limits or other questions with trig functions, it may be best to get some of the identities on hand or to try and remember a couple. Just going tobe what youre forced to do. Well, what I type above is an identity, and we can use it to replace the -sin^2(x).

So if I subtract sin^2(x) from both sides in the identity, we get:

\[\cos^{2}x = 1 -\sin^{2}x\]then subtracting 1 from both sides I have:
\[\cos^{2}x - 1 = -\sin^{2}x\]
So that means I can replace -sin^2(x) with cos^2(x) - 1. that gives me now:

\[\frac{ \cos^{2}x -1 }{ cosx+1 }+cosx+4\]Now once again, you can factor that cos^2(x)-1. Do you know how to ?

Answer 27

cos x-1 cos x+1

Answer 28

Awesome. So now you canceled out the cosx + 1 everything, thats what we needed to get accomplished. So we're left with:
\[cosx - 1 + cosx + 4\]
\[\lim_{x \rightarrow \pi}2cosx + 3\]ALl I did was combine like terms at this step. SO now just plug in pi for x and solve.

Answer 29

Thanks so much.. I did not take trig ..

Answer 30

Oh O.o Youre in calculus, right?

Answer 31

yes..

Answer 32

Yeah, thats definitely quite odd to have not done trig and be in calculus. It might be a big problem, depending on the teacher and how crazy they want to go into trig problems. You will have to know some basic trig things, theres no way to get around that, though.

Answer 33

i took pre cal.. so the answer will be 4.99

Answer 34

Usually trig is a part of pre-cal. And the answer is actually just 1. Is your calculator in degrees mode or radians? It needs to be in radians.

Answer 35

deg

Answer 36

Yeah, thats why you got a wrong answer, you need radians.

Answer 37

that answer isn't right on my homework submission thing

Answer 38

The 1?

Answer 39

yes

Answer 40

If I have the question you posted correct, then the limit is 1. Unless the problem is written wrong or the question wnats something else.

Answer 41

\[\lim_{x \rightarrow \pi}\frac{ \cos2x + 5cosx + 4 }{ cosx + 1 }= 1\]
Unless the problem is wrong then yeah, limit is 1, not sure what would be wrong about it :/

Answer 42

idk.. that's the problem

Answer 43

Got a way to screenshot it?

Answer 44

Thats the problem, you have it written liek cos2x, not cos^2(x)

Answer 45

oh.. sorry

Answer 46

Well, that just means the limit was a little less work. So we factor the top. We already facotred that same thing before, so we can just skip ahead and say we have:
\[\frac{ (cosx+1)(cosx+4) }{ cosx+1 }\implies cosx+4\]
Now plug in pi for x.

Answer 47

3

Answer 48

do you mind helping me with about 3 more problems?

Answer 49

I can try. Just know that some of the trig stuff you will have to learn about. But lets see what ya got.

Answer 50

Well, if you draw your graph left to right, at what point are you forced to pick up your pencil in order to graph the next point?

Answer 51

4,-3

Answer 52

1,3

Answer 53

for the on labeled 3p i thought i could cancel the like terms but it will give me zero in both the numerator and denominator

Answer 54

Well, for the graph one, youre mainly looking at the points on the graph where there are circle holes. As for #3, they do that on purpose. The whole idea is you need to know how to manipulate it so you DONT get 0 on bottom.

Answer 55

im not sure how to do this?

Answer 56

Well, its about knowing algebra well and trig. As for the 2nd problem, just put the coordinates of the spots where there are holes. A graph is discontinuous at any spot where you have to pick up your pencil in order to plot a point.

Answer 57

I did that one.. im on the 3p problem..

Answer 58

Alrighty. But yeah, you always have to manipulat ethese things to try and get the 0 out of the bottom. Alrighty, just was checking it out real fast to see how bad it might be. Okay, would you know how to make the top two fractions in your problem into one fraction?

Answer 59

would you multiply by 1/2

Answer 60

or by sin

Answer 61

You might be able to. I just kinda look at it how I think would be easiest. So you need to make a common denominator between 2 and 2+ sinx

Answer 62

would it be 2+sin

Answer 63

I think you need to multiply the two and say its 2(2+sin). So when you finally cimbine the two fractions, the -1 on top of the right fraction must be multiplied by the 2+sinx, because that was the part of the common denominator that it was missing. As for the left fraction, you multiply the 1 on top by 2, since that is a part of the common denominator that was missing.

Answer 64

can you write that in the numerical way?

Answer 65

Combine the two now:

\[\frac{ 2-(2+sinx) }{ 2(2+sinx) }=\frac{ 2-2-sinx }{ 2(2+sinx) }\implies \frac{ -sinx }{ 2(2+sinx) }\]

Kinda see at all what I did?