the question is

f(x)=(2x+sin(x))/(2x)

as x approaches 0

 

split the equation and I get

f(x)=(2x)/(2x) + sin(x)/(2x)

(2x)/(2x)=1?

sin(x)/(2x)=(1/2)? (since sin(x)/x=1)

 

add the 2 together and I get the answer 3/2

 

now this is where the trouble starts, when I graph this function, as x approaches 0 from possible and negative side. the y value is 1.00873

 

so which answer is correct? the 3/2 from algebra or 1.00873 from graphing?

I probally did something wrong somewhere, help me

Your algebra is correct, even according to my plot.

Perhaps you made a wrong graph?

I checked, I entered the equation correctly. Graphed it on my Ti-89 with pretty print on so it was easy to check if I entered correctly. this is puzzling

My plot with both Maple and Graphmatica show clearly that f(0) = 3/2.

damn it, I found the problem, I was in degree mode. guess any equation with trig fuction in it has to be graphed on radian mode. thanks

I was about to suggest the very same thing, but obviously you got there before me

at the risk of sounding INCREDIBLY stupid...

 

since when was sin(x)/x=1?

 

Because then you mulitply both sides by x, and then sin(x)=x!!!

now this is where the trouble starts, when I graph this function, as x approaches 0 from possible and negative side. the y value is 1.00873

 

so which answer is correct? the 3/2 from algebra or 1.00873 from graphing?

I probally did something wrong somewhere, help me

I think and I hope I know what went wrong.

You didn't turn the angle into radian, right?

at the risk of sounding INCREDIBLY stupid...

 

since when was sin(x)/x=1?

 

Because then you mulitply both sides by x, and then sin(x)=x!!!

The case is dealing with limit./

ok, so explain this to me... when manipulating a formula for a limit, you can say sin(x)/x=1? becase is you take the limit with sin(x)/x, you get sin(0)/0. sin(0)=0, so it's 0/0. But when did we learn that you use different rules when manipulating the formula for taking a limit?

It's a valid question, BobbyJoeCool and important to follow the solution to this fairly small problem. You solve lim x -> 0 sin(x) /x type of question using L'Hopital's rule (sp?). In other words,

limit of f(x)/g(x) = limit of f'(x)/g'(x)

which is really useful for cases when you get indeterminates like 0/0 or inf/inf using normal rules. So

lim x -> 0 sin(x)/x = sin(0)/0 = 0/0 = indeterminate, so we differentiate the top and bottom and get

lim x -> 0 cos(x)/1 = cos(0)/1 = 1/1 = 1. And therefore, by L'Hopital's rule,

lim x -> 0 sin(x)/x = 1.

 

Remember your calculus!

I'm in calc 1... have we haven't "learned" that rule yet (although I know it because my teacher decided to give us a problem I have posted in another thread somewhere). But that makes sence now... thanks!

I am currently studying some simple limit and function.

If I didn't browse this web, I didn't know I've entered the Calculus branch.

The education system does not fascinate us by the beauty of mathematics.

in my calculus class, we learned lim x->0 of sin(x) / x is 1 and that lim x -> 0 of cos(x) + 1 / x is 0, however we don't tackle l'hoptal (sp)'s rule until second semester.

It's a valid question' date=' BobbyJoeCool and important to follow the solution to this fairly small problem. You solve lim x -> 0 sin(x) /x type of question using L'Hopital's rule (sp?). In other words,

limit of f(x)/g(x) = limit of f'(x)/g'(x)

which is really useful for cases when you get indeterminates like 0/0 or inf/inf using normal rules. So

lim x -> 0 sin(x)/x = sin(0)/0 = 0/0 = indeterminate, so we differentiate the top and bottom and get

lim x -> 0 cos(x)/1 = cos(0)/1 = 1/1 = 1. And therefore, by L'Hopital's rule,

lim x -> 0 sin(x)/x = 1.

 

Remember your calculus! [/quote']

 

 

can anyone give the steps to find the derivative of sin(x)

(sin(x+h)-sin(x))/h

what then? use sum double angle formula?

Check http://www.math2.org/math/derivatives/more/trig.htm

the derivitive of sin(x)=cos(x)...

Yes, but he was more interested in how you actually show that using limits

oh... right. Limit derivaitve. that would be moderatly difficult...

can anyone give the steps to find the derivative of sin(x)

(sin(x+h)-sin(x))/h

what then? use sum double angle formula?

 

I tried to figure this out myself just today, couldnt think of how to do it so i just plotted / guess and checked and its cos(x)

 

someone else would have to explain why its cos(x)

sorry, double post... the one below is correct (pressed wrong button). If a mod would delete this, that would be nice of them!

I tried to figure this out myself just today' date=' couldnt think of how to do it so i just plotted / guess and checked and its cos(x)

 

someone else would have to explain why its cos(x)[/quote']

 

from the site quoted above by TD...

 

[math]\lim_{h \to 0} \frac{sin(x+h)-sin(x)}{h}[/math]

 

double angle formula

[math]\lim_{h \to 0} \frac{sin(x)cos(h)+sin(h)cos(x)-sin(x)}{h}[/math]

 

spilit the limits

[math]\lim_{h \to 0} \frac{sin(x)cos(h)-sin(x)}{h}+\lim_{h \to 0} \frac{sin(h)cos(x)}{h}[/math]

 

take variables out (limit will still be the same since when h->0, x might as well be a constant)

[math]sin(x) \lim_{h \to 0} \frac{cos(h)-1}{h}+cos(x) \lim_{h \to 0} \frac{sin(h)}{h}[/math]

 

multiply numerator and demoninator of the first one by (cos(x)+1) so you can get cos^2-1 in the numerator which can later be manipulated becaus sin^2+cos^2=1

[math]sin(x) \lim_{h \to 0} \frac{cos^2(h)-1}{h \cdot (cos(h)+1)}+cos(x) \lim_{h \to 0} \frac{sin(h)}{h}[/math]

 

manipulates, explained above.

[math]sin(x) \lim_{h \to 0} \frac{-sin^2(h)}{h \cdot (cos(h)+1)}+cos(x) \lim_{h \to 0} \frac{sin(h)}{h}[/math]

 

uses a proporty of limits to seporate a bunch of things so you end up with a bunch of do-able limits... (lim a*b= lima * lim b)

[math]sin(x) \left( \lim_{h \to 0} \frac{-sin(h)}{cos(h)+1} \right) \cdot \left( \lim_{h \to 0} \frac{sin(h)}{h} \right) \cdot \left( \lim_{h \to 0} \frac{1}{cos(h)+1} \right) + cos(x) \lim_{h \to 0} \frac{sin(h)}{h}[/math]

 

[math]sin(x) \cdot 0 \cdot 1 \cdot \tfrac{1}{2} + cos(x) \cdot 1=cos(x)[/math]

in the last step, how did you get [math] ( \lim_{h \to 0} \frac{1}{cos(h)+1}) [/math] ? I can see how -sin^2 split into sin(h) and -sin(h).

That factor shouldn't be there I believe.

It shouldn't... I was both copying and doing it myself...it really should be

 

[math]sin(x) \left( \lim_{h \to 0} -sin(h) \right) \cdot \left( \lim_{h \to 0} \frac{sin(h)}{h} \right) \cdot \left( \lim_{h \to 0} \frac{1}{cos(h)+1} \right) + cos(x) \lim_{h \to 0} \frac{sin(h)}{h}[/math]