Evaluate Limit: limx→0 (sinx/x)=1?

[charmedbeauty]

limx-->0 (sinx/x)=1?

Homework Statement

Hi I have the question evaluate the limit going to zero of

(cos(x)-1)/sin(x²)

Homework Equations

The Attempt at a Solution

lim x→0 (cos(x)-1)/sin(x²)

goes to 0/0 by Lhoptal's

f'(x)/g'(x)

= (limx→0 sinx/x)(limx→0 -1/ 2cos(x²))

but the answer I have says that (limx→0 sinx/x) = 1

how is this so

I know the other term goes to -1/2 but yeah I'm really confused about the sin term going to 1 is this a mistake or is there something i don't know?

thanks.

 

[Dickfore]

Hint:
<br /> \cos x - 1 = -2 \sin^2 \left( \frac{x}{2} \right)<br />
Divide the numerator and denominator by x^2, and use your limit.

 

[charmedbeauty]

Hint:
<br /> \cos x - 1 = -2 \sin^2 \left( \frac{x}{2} \right)<br />
Divide the numerator and denominator by x^2, and use your limit.

do you mean cos²x-1=-2sin²(x/2)?

if i divide numerator and denominator by x²

(sinx × x²)/x³

I still get 0 out of it when I plug 0 into anything times by sin i get 0.

do I take l'hospital's again?

of the individual sin part of the function? can you partially diff and apply l'hospital's?

 

[Dickfore]

do you mean cos²x-1=-2sin²(x/2)?

No, because:
a) it is an incorrect trigonometric identity;
b) it is not useful for your problem, since the left-hand side is not what you have in your limit.

I meant exactly what I wrote:
<br /> \cos x - 1 = -2 \, \sin^2 \left( \frac{x}{2} \right)<br />

if i divide numerator and denominator by x²

(sinx × x²)/x³

I don't know how you got this.

 

[clamtrox]

but the answer I have says that (limx→0 sinx/x) = 1

how is this so

What happens if you try L'Hopital again?

 

[HallsofIvy]

I think I would be inclined to write it as
\frac{cos(x)- 1}{sin(x^2)}= \frac{x^2}{sin(x^2)}\frac{cos(x)- 1}{x^2}

It's easy to see that sin(x^2)/x^2 goes to 1. Using L'Hopital's rule on \frac{cos(x)- 1}{x^2} leads to \frac{-sin(x)}{2x} which has limit -2.

 

[charmedbeauty]

I think I would be inclined to write it as
\frac{cos(x)- 1}{sin(x^2)}= \frac{x^2}{sin(x^2)}\frac{cos(x)- 1}{x^2}

It's easy to see that sin(x^2)/x^2 goes to 1. Using L'Hopital's rule on \frac{cos(x)- 1}{x^2} leads to \frac{-sin(x)}{2x} which has limit -2.

Ok, I can see how you get that but the part I'm confused with is that you have a cos factor in the numerator and a sin factor in the denominator. every time L'hopital's gets used these just swap from sin to cos visca versa.

and cos 0 =1 but sin0=0

thats how I don't understand how the zero term vanishes?

can you use l'hospital's rule on just part of a fraction?

because it looks like every time I differentiate tops and bottoms I have a sin and a cos factor, when i really only can have cos factors.?

 

[Muphrid]

This is much more easily solved with Taylor expansion.

\cos x - 1 = \left(1 - \frac{x^2}{2} + \ldots{}\right) - 1 = -\frac{x^2}{2} + \ldots{}

And for the denominator,

\sin x^2 = x^2 - \frac{x^6}{6} + \ldots{}

So the limit can be written as

\lim_{x \to 0} \frac{\cos x - 1}{\sin x^2} = \lim_{x\to 0} \frac{-x^2/2 + \ldots}{x^2 - x^6/6 + \ldots{}}

The limit is as x \to 0, so throw away all but the lowest power term in both the numerator and denominator, and you clearly get

\lim_{x \to 0} \frac{-x^2/2}{x^2} = -\frac{1}{2}

Now, you originally asked how it could be that,

\lim_{x \to 0} \frac{\sin x}{x} = 1

Again, this is easily solved by Taylor expansion.

\lim_{x \to 0} \frac{x - x^3/6 + \ldots}{x} = \lim_{x\to 0} \frac{x}{x} = 1

Or by l'Hopital's rule,

\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1

 

[HallsofIvy]

Ok, I can see how you get that but the part I'm confused with is that you have a cos factor in the numerator and a sin factor in the denominator. every time L'hopital's gets used these just swap from sin to cos visca versa.

and cos 0 =1 but sin0=0

thats how I don't understand how the zero term vanishes?

I'm not sure what you mean by that. "vanishes" means "is 0".

can you use l'hospital's rule on just part of a fraction?

You understand what L'Hopital's rule is, don't you?
\lim_{x\to a}\frac{f(x)}{g(x)}= \frac{\lim_{x\to a}f&#039;(x)}{\lim_{x\to a} g&#039;(x)}
provided those limits exist. I don't know what you mean by "part of a fraction". If you can write a fraction as a sum or product, then you can use \lim_{x\to a} (f+g)(x)= \lim_{x\to a}f(x)+ \lim_{x\to a}g(x) and \lim_{x\to a}fg(x)= \left(\lim_{x\to a}f(x)\right)\left(\lim_{x\to a}g(x)\right).

because it looks like every time I differentiate tops and bottoms I have a sin and a cos factor, when i really only can have cos factors.?

Again, I really don't know what you mean by this. Why would you think "I really only can have cos factors"?

 

[vela]

This is much more easily solved with Taylor expansion.

This method likely requires material the OP doesn't yet know considering he or she is still learning how to evaluate basic limits.

 

[charmedbeauty]

This method likely requires material the OP doesn't yet know considering he or she is still learning how to evaluate basic limits.

yep.

 

[charmedbeauty]

Again, I really don't know what you mean by this. Why would you think "I really only can have cos factors"?

ok so its ok to have...

limx→0 (-1/2cos(x²)limx→0(cos(x)/1)

= -1/2 (1) =-1/2

??

 

[vela]

Applying L'Hopital's rule to what you started with, you got
$$\lim_{x \to 0} \frac{\cos x - 1}{\sin x^2} = \lim_{x \to 0} \frac{-\sin x}{2x\cos x^2}$$ I think what you're missing is this next step. What happened was they broke the single limit into a product of two limits:
$$\lim_{x \to 0} \frac{-\sin x}{2x\cos x^2} = \left(\lim_{x \to 0} \frac{\sin x}{x}\right)\left(\lim_{x \to 0} \frac{-1}{2\cos x^2}\right)$$ which you can do if the limits exist. So while it looked like L'Hopital's rule was only being applied to part of the fraction, namely, the sin x/x part, it really wasn't. It's just that the original limit was split into two separate limits. Now the first limit you can evaluate using L'Hopital's rule again.

 

[charmedbeauty]

Applying L'Hopital's rule to what you started with, you got
$$\lim_{x \to 0} \frac{\cos x - 1}{\sin x^2} = \lim_{x \to 0} \frac{-\sin x}{2x\cos x^2}$$ I think what you're missing is this next step. What happened was they broke the single limit into a product of two limits:
$$\lim_{x \to 0} \frac{-\sin x}{2x\cos x^2} = \left(\lim_{x \to 0} \frac{\sin x}{x}\right)\left(\lim_{x \to 0} \frac{-1}{2\cos x^2}\right)$$ which you can do if the limits exist. So while it looked like L'Hopital's rule was only being applied to part of the fraction, namely, the sin x/x part, it really wasn't. It's just that the original limit was split into two separate limits. Now the first limit you can evaluate using L'Hopital's rule again.

Thanks vela that completely cleared it up.