Question about substitution in limits?

[Kratos321]

okay so I just have a question about using substitution when solving limits.

Say I have the function sinx/x, if i want to find the limit when x--> 0 using approximation ( i know how to prove it with the pinching theorem by the way). So if I substitute 0.0000000001 rad into the function i get something close to 1. However if i substitute that many DEGREES I am waaaaaaaayyyyy off. I know a rad is a lot more than a degree but if i draw the graphs for both wouldn't i see that the limit is 1 for both the rad graf and the degree. (This isn't an actual question where i have to use substitution but i was just experimenting and am kinda wondering...).
any explanations would be appreciated.

thank you.

 

[Bacle]

Are you sure you're expressing _both_ sinx and x in degrees when you do your calculation?

 

[mathman]

sinx/x -> 1, only when the argument of the sin is in radians. Mathematically, the argument for all trig functions (when being considered as functions) has to be in radians.

 

[Bacle]

What I meant is you can look at working with degrees as just rescaling by 2Pi, i.e., start with 360=2Pi , and rescale any angle . If x->0 , then x':=2Pix/360 is a rescaling, and
sinx'/x' also goes to 1 .

 

[Kratos321]

yeah that's what i figured but is there a reason or explanation for it?

 

[Char. Limit]

yeah that's what i figured but is there a reason or explanation for it?

sin(x_r) and sin(x_d) (where x_r and x_d represent "radian x" and "degree x") are different functions, that's all. Namely, sin(x_d) = sin(pi/180 x_r). And so reasonably, the two functions would approach zero at different "speeds".

 

[Bacle]

Re: Different Speeds (My 'quote' is not working)

Char Limit: the rescaling here makes no difference at the end, e.g., try

L'Hopital:

Sin(kx)/kx --> kCos(kx)/k= Cos(kx) , goes to 1 as x->0 , since k<oo ; just use, e.g

Taylor's Thm.

 

[Char. Limit]

Re: Different Speeds (My 'quote' is not working)

Char Limit: the rescaling here makes no difference at the end, e.g., try

L'Hopital:

Sin(kx)/kx --> kCos(kx)/k= Cos(kx) , goes to 1 as x->0 , since k<oo ; just use, e.g

Taylor's Thm.

Incorrect. It makes no difference if you scale x similarly, but just using the "degree sine" function without scaling x WILL make it change. We know this:

\lim_{x \to 0} \frac{sin_r(x)}{x} = 0

where sin_r(x) represents the "radian sine" function, which is the usual sine function. For reference, sin_d(x), representing the "degree sine" function, is equal to sin_r(pi/180 x). So therefore:

\lim_{x \to 0} \frac{sin_d(x)}{x} = \frac{sin_r\left(\frac{\pi}{180} x\right)}{x} = \frac{\pi}{180}

You see why, I hope.

 

[Bacle]

I only referred to a scaling, not a change into degrees; still, sinx is defined in R, i.e.,

for all reals.

And what's with the 'I Hope' ?

 

[Bacle]

And, BTW, I clearly stated that the rescaling should be done both with the argument

and with x, and my argument reflects that ; I think you misunderstood/misinterpreted

my answer.