prove limit of sin(x) as x approaches zero = 0

[Maddy_Math]

Can we prove:

[ limit of sin(x) as x ---> 0 = 0 ]

using delta epsilon notation. I tried to do it but stuck at first step:

|sin(x) - 0| = |sin(x)| what next how to lead it to a value where it will be less than some epsilon??? for some delta > |x - 0| > 0

 

[HallsofIvy]

So you want to prove that sin(x) is continuous at x= 0? How you do that depends heavily upon how you have defined "sin(x)". What definition are you using?

 

[pka]

[ limit of sin(x) as x ---> 0 = 0 ]
using delta epsilon notation. I tried to do it but stuck at first step:
|sin(x) - 0| = |sin(x)| what next how to lead it to a value where it will be less than some epsilon??

Can you show that $\displaystyle |\sin(x)|\le |x|~?$

 

[Maddy_Math]

So you want to prove that sin(x) is continuous at x= 0? How you do that depends heavily upon how you have defined "sin(x)". What definition are you using?

I guess sin(x) = y/r simple

 

[Maddy_Math]

Can you show that $\displaystyle |\sin(x)|\le |x|~?$

I guess yes using MVT:

first sin(x) is less than x for all values of x > 1 because sin(x) is bounded by interval [-1,1]

now for interval [0,1] also let c belong to [0,1]

by MVT:

{[ f(b) - f(a)] / [ b - a ]} = f'(c)

by choosing f(x) = sin(x)

[ sin(x) - sin(0) ] / [ x - 0 ] = sin'(c) <===> [ sin(x) / x ] = cos(c) ]

===> [ sin(x) / x ] <= 1 because c belongs to [0,1] and so cos(over such a c) is always gonna be <= 1
<===> sin(x) <= x

 

[pka]

I guess yes using MVT:
first sin(x) is less than x for all values of x > 1 because sin(x) is bounded by interval [-1,1]
now for interval [0,1] also let c belong to [0,1]
by MVT: {[ f(b) - f(a)] / [ b - a ]} = f'(c) YES!
by choosing f(x) = sin(x)

Now let $\displaystyle \delta=\varepsilon>0$.

 

[Maddy_Math]

Now let $\displaystyle \delta=\varepsilon>0$.

lolx I'm done but I have a confusion can't I prove any limit to exist this way even if it doesn't exist???

 

[pka]

I have a confusion can't I prove any limit to exist this way even if it doesn't exist???

Well that limit does exist. So what is your point?

 

[stapel]

...can't I prove any limit to exist this way even if it doesn't exist?

Why do you think the limit does not exist? Please be specific. Thank you!

 

[Maddy_Math]

for example $\displaystyle \lim \limits_{x \to 1} \frac{\text{x+1}}{\text{x-1}}$ does not exist at x = 1 but can't I use delta epsilon way to prove that it exists at 1 like

$\displaystyle \mid \frac{\text{x+1}}{\text{x-1}} \text{- L} \mid < \epsilon \text{ Whenever } \mid \text{x - 1} \mid < \delta$

can't I prove this to exist now???

 

[stapel]

for example $\displaystyle \displaystyle{\lim_{x\, \to\, 1}\, \frac{x\,+\,1}{x\,-\,1}}$ does not exist at x = 1

Oh. Have you switched from the original sine question to something else?

can't I use delta epsilon way to prove that it exists at 1 like

\(\displaystyle \left|\, \dfrac{x\,+\,1}{x\,-\,1}\, -\, L\, \right|\, <\, \epsilon \,\mbox{ whenever }\, \left|\, x\, -\, 1\, \right| \, < \,\delta \)

can't I prove this to exist now???

Since clearly the function is not defined at x = 1, I would be interested in seeing all of the steps in your "proof" that it does actually exist at x = 1. :shock: Please start with the value you have determined for $\displaystyle L.$

 

[Maddy_Math]

Oh. Have you switched from the original sine question to something else?


Since clearly the function is not defined at x = 1, I would be interested in seeing all of the steps in your "proof" that it does actually exist at x = 1. :shock: Please start with the value you have determined for $\displaystyle L.$

lolx I tried but got stuck at $\displaystyle \mid \frac{2}{x-1} \mid$

thanks stapel, now it makes sense