limit question, confusing

[caseclosed]

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

[TD]

Your algebra is correct, even according to my plot.

Perhaps you made a wrong graph?

[caseclosed]

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

[PhDP]

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

[caseclosed]

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

[Dave]

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

[BobbyJoeCool]

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!!!

[Primarygun]

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./

[BobbyJoeCool]

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?

[DoorNumber1]

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!

[BobbyJoeCool]

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!

[Primarygun]

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.

[danielS]

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.

[caseclosed]

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?

[TD]

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

[BobbyJoeCool]

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

[Dave]

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

[BobbyJoeCool]

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

[CanadaAotS]

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)

[BobbyJoeCool]

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

[BobbyJoeCool]

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]

[caseclosed]

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).

[TD]

That factor shouldn't be there I believe.

[BobbyJoeCool]

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]