Question

Limits HELP!!

Answer 1

@Fifciol @ash2326

Answer 2

No Idea wat that is!!

Answer 3

I mean I m8 be knowing the rule...But dunno if its L'Hopital's!!
I heard Hopital's rule in differentiation!!

Answer 4

And??

Answer 5

hw 2 apply it here??

Answer 6

Ummmmm........Guys???

Answer 7

\[\frac{ x }{ x+1 }\le \frac{ 1+x+sinx }{ x+cosx } \le \frac{ 2+x }{ x-1 }\]
using squeeze theorem you can easy check that this limit is 1

Answer 8

Is this correct???

Answer 9

@Fifciol Is it r8
????

Answer 10

I dunno

Answer 11

one from your eyeballs

Answer 12

Sry??

Answer 13

sine and cosine are stuck between -1 and 1, they are irrelevant to this problem

Answer 14

So just ignore them??
Then I get 0/0 +1

Answer 15

you have \(\frac{x}{x}=1\)

Answer 16

this is clear right? imagine even if \(x=10^3=1000\) not really that big of a number
you have basically 1000/1000= 1 and can only be off by at most 2 on the top and 1 on the bottom

Answer 17

but if you want you can certainly use @Fifciol squeeze theorem above

Answer 18

I have no idea of sqeeze theo1!!:(

Answer 19

But @satellite73 I learnt 0/0 is undefined!!what say to that??

Answer 20

@Loser66 sinx/x=1 when x tends to 0 and not infinity:(

Answer 21

lol

Answer 22

@Fifciol @satellite73 @Loser66 CAn u guys suggest a good link to learn squeeze theorem??

Answer 23

it's very simple , you bound left side by something that you now that is less or equal then mid one in this case sinx and cosx have values in interval <-1;1> so the left side to be always less or equal I want numerator to be the smallest ( so sinx must be -1) and denominator to be the largest ( cosx =1) and analogically you do it with right side.
If the limit of left side is the same as right side the limit of mid one expression must be equal to that value of limit of each side that's what the squeeze theorem is all about

Answer 24

Like @satellite73 mentioned, you can use the fact that \(\dfrac{x}{x}=1\) for nonzero \(x\).
\[\lim_{x\to\infty}\frac{x+\sin x+1}{x+\cos x}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}\\
\lim_{x\to\infty}\frac{1+\frac{\sin x}{x}+\frac{1}{x}}{1+\frac{\cos x}{x}}
\]
Now you have \(\dfrac{\sin x}{x},\dfrac{\cos x}{x},\dfrac{1}{x}\to0\) as \(x\to\infty\), leaving you with
\[\lim_{x\to\infty}1\]

Answer 25

It looks like what you did, with the exception of using an intermediate substitution along the way.

Answer 26

Tnx @SithsAndGiggles But isnt sinx/x where x tends to infinity 0??

Answer 27

Its 1 when x tends to 0!!

Answer 28

Yeah, didn't I say that?

Answer 29

Did U??sry ...cudnt getcha!!

Answer 30

\[\large \dfrac{\sin x}{x},\dfrac{\cos x}{x},\dfrac{1}{x}\to0~~\text{as}~~x\to\infty\]

Answer 31

@Fifciol Tnx for the help on sqeeeeeeeze theo!!

Answer 32

Now I got it....

Answer 33

I wouldn't have used it if I had thought that sinx/x goes to zero when x goes to infinity, but I've reminded myself squeeze theory tho :)

Answer 34

Love the phrase sqeeeeeeeeeeeeeeze theorem!
Anyways Cheers!!