Question

would the limit for lim x-->0
e^x sin x
be equal to 0??

Answer 1

or 1.....?

Answer 2

Just 0. Because yeah, e^0 = 1, but sin(0) = 0, so 0 *1 = 0

Answer 3

thought so :) so you can basically just plug in the number to the equation ?

Answer 4

lol sorry for dumb questions, i'm not very bright :(

Answer 5

Yep. As long as you dont get an undefined answer. sin(0)e^0 isnt undefined, so I could care less if it still spits out 0 xD

Answer 6

what do you do when it asks about ints?? for ex
lim x--> 7/2-
int (2x-1) ? :o

Answer 7

int?

Answer 8

that's the hw problem >__< D:

Answer 9

what does that even mean? lol

Answer 10

Not sure what int would be, but wouldnt it just be 6? xD

Answer 11

that would be my guess as well, but i graphed it and i get like 4.5 when i trace it

Answer 12

?D:

Answer 13

In order for you to graph it wouldnt you have to know what int is? xD

Answer 14

ok, we'll just go with 6(: jaja

Answer 15

okay.

Answer 16

QUESTIONNNN

Answer 17

lol

Answer 18

when your looking at limits approaching infinity

Answer 19

do you look at what is going to left/ right of the graph or whats going up and down??

Answer 20

Well, limits are literally "what is the y-value at x = ?" So youre checking left and right to observe up and down. So for infinite limits, you check values appraoching from the left and appraoching from theright and see if they both shoot off to infinity or both shoot off to negative infinity. If they go in opposite directions then the limit doesnt exist.

Answer 21

so correct me if i'm wrong but x-->infinity
(x+sin(x))/(x+cos(x)) is DNE?

Answer 22

In this case, no actually.

Answer 23

??:( whyyyyyyyy? lol

Answer 24

Im checking xD I just punched into my calculator real fast without checking it, but Im working on confirming.

Answer 25

Okay, yeah, we have to do it this way. Have you see the method of dividing everything by the highest power in the denominator?

Answer 26

.__. no lol

Answer 27

Ah x_x Okay, so a normal trig when a function has x -> infinity or negative infinity is to find the highest power in the denominator and divide all terms in the function by that power. So in the denominator, all we have is a 1st power of x. So we want to divide all terms by x to start:

\[\frac{ \frac{ x }{ x }+\frac{ sinx }{ x } }{ \frac{ x }{ x }+\frac{ cosx }{ x} }\]which simplifies to:

\[\frac{ 1+\frac{ sinx }{ x } }{ 1+\frac{ cosx }{ x } } \]Now we have to take x to infinity. Now, two things to mention. In general, if we have a function like:

\[\lim_{x \rightarrow \infty}\frac{ x^{a} }{ x^{b} }\] and a > b, then the limit is infinity. If b > a, then the limit is 0. So thats something to be aware of. However, when you have trig functions you have to be a bit more careful. You just just say it goes to 0 because theres an x in the denominator. Now, some textbooks consider these identities, some make you show them, but:

\[\lim_{x \rightarrow \infty}\frac{ sinx }{ x }= 0\]
\[\lim_{x \rightarrow 0}\frac{ cosx }{ x }=0\]Now, if thse werent identities and you had to prove them, youd have to prove them with squeeze theorem. Basically, the squeeze theorem requires you to show, using inequalities, that the left inequality and the right inequality are the same as limit goes to infinity. So this is how itd bedone:

\[-1 \le sinx \le 1\]The range of the sine function is -1 to 1, so this inequality is valid. But we have sinx/x, so lets divide everything by x
\[\frac{ -1 }{ x } \le \frac{ sinx }{ x } \le \frac{ 1 }{ x } \]Now as I said, the goal is to have the left inequality and the right inequality to have the same limit as x goes to infinity. Well, as was just mentioned above, if the power of x is higher in the denominator, which it is in -1/x and 1/x, then limit is 0. So basically, since the left inequality is 0 and the right inequality is 0, sinx/x also has to be 0. This exact same thing can be done with cosx/x.

Either way, we end up with:

\[\frac{ 1+0 }{ 1+0 }= 1\]

I know thats a lot of info, so feel free to ask anything lol.

Answer 28

YOU. ARE. AWESOME. omg. thank you so much!!!!

Answer 29

i get it :)

Answer 30

Okay, cool xD Just yeah, if you ever come up with a trig function with these limits, it may require yout o try and mess around with the squeeze theorem, especially with ones that go to infinity.

Answer 31

got it. :)

Answer 32

seriosuly thanks so much for that

Answer 33

i would give you a thousand medal just for that long long explanation lol

Answer 34

Yah, np, lol. You wont have to be doing much more with limits once you finish that chapter, but squeeze theorem and knowing about dividing everything by the highest power in the DENOMINATOR or checking limits left and right. All the techniques you can use xD

Answer 35

when being asked question like at what point is the discontinuity, you're looking for y value right?(:

Answer 36

lol you're a life saver:D

Answer 37

Well, depends. It may be wanting the whole coordinate. Its either going to be the whole coordinate or just the x-coordinate, theyre not going to be asking the y-value if it just asks wheres the discontinuity.

Answer 38

Most like x-coordinate really, since an asymptote we would just name as x = ?

Answer 39

what does it mean when it says find a power function end behavior model? :o

Answer 40

That sounds weird, lol. Power function? O.o

Answer 41

yeahh .__. lol

Answer 42

i have no clue lol

Answer 43

Yeah, no idea. End behavior is fine on its own, direction the graphs starts from and the direction the graph ends towards. Power function, dunno.

Answer 44

so when it says find a right end behavior model and a left end behavior model of a function you just describe, draw it.........? .__.

Answer 45

lol

Answer 46

f(x)=ln abso value(x) +sin x

Answer 47

JK don't have to study that one :)

Answer 48

ok, i should REALLY stop bugging you now lol!

Answer 49

you've been tooooons of help thanks yo! (:

Answer 50

seriously. you are awesome!

Answer 51

Lol, its fine xD Sorry I didnt respond immediately xD

Answer 52

no seriously thank you so much :)

Answer 53

if it's fine i have just one last question for you and then we can part ways :) lol

Answer 54

Go for it.

Answer 55

what would the graph look like if lim f(x) did not exist and
lim f(x)=f(2)=3 ? as x->2

Answer 56

Sounds like its basically saying:
\[\lim_{x \rightarrow 2}f(x) = 3\] But its just written in an odd way O.o

Answer 57

hm, so it'd be 3?

Answer 58

i mean sorry lol

Answer 59

that was completely wrong

Answer 60

wouldn't it be a discont graph that has jump disc?

Answer 61

I guess im not sure how else to interpret it. Is it written exactly like that?

Answer 62

yeaahh

Answer 63

Yah, itd have to be a jump discontinuity it looks like. Only way for f(2) to be defined but not have a limit.

Answer 64

ok, i'll just ask my teach to see if my sketch is correct :)

Answer 65

I think the qustion is too vague to have a perfect sketch, just as long as you have some sort of graph that has a coordinate (2,3) that exists but then jumps to some other y-value ax x = 2 xD

Answer 66

alrighty, thank you @Psymon for the very useful help!

Answer 67

np ^_^

Answer 68

yeah :) i got that lol

Answer 69

THANK YOU.

Answer 70

bye! :)

Answer 71

bye :3