Thread: trigonometric limits

1. trigonometric limits

Evaluate the following limit:

${\lim_{x \to \0} \frac{{\sin }^{ 2} (x)}{{x}^{2} }$

Ok i really don't know what to do

so were trying to get $\sin {x}^{2 }$

2. Use the result

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

3. Originally Posted by purplec16
Evaluate the following limit:

${\lim_{x \to \0} \frac{{\sin }^{ 2} (x)}{{x}^{2} }$

Ok i really don't know what to do

so were trying to get $\sin {x}^{2 }$
Another way to think about this is in terms of the derivative.

Let

$f(x)=\frac{\sin(x)}{x}=\frac{\sin(x)-0}{x-0}$

so the limits is

$\lim_{x \to 0}[f(x)]^2=\left( \lim_{x \to 0} f(x)\right)\left( \lim_{x \to 0} f(x)\right)$

but each factor is just the definition of the derivative evaluates at zero

but

$\frac{d}{dx}\sin(x) \bigg|_{x=0}=1$

$\lim_{x \to 0}[f(x)]^2=\left( \lim_{x \to 0} f(x)\right)\left( \lim_{x \to 0} f(x)\right)=1\cdot 1=1$

4. Originally Posted by pickslides
Use the result

$\displaystyle \lim_{x\to 0 } \frac{\sin x}{x} = 1$
I know that but how do I make the numerator x^2

5. $\displaystyle \frac{\sin^2 x}{x^2} = \frac{\sin x}{x}\times \frac{\sin x}{x}$

6. wow... so true...i can't believe i didnt see that -.-

thus the answer is 1

7. Originally Posted by TheEmptySet
Another way to think about this is in terms of the derivative.

Let

$f(x)=\frac{\sin(x)}{x}=\frac{\sin(x)-0}{x-0}$

so the limits is

$\lim_{x \to 0}[f(x)]^2=\left( \lim_{x \to 0} f(x)\right)\left( \lim_{x \to 0} f(x)\right)$

but each factor is just the definition of the derivative evaluates at zero

but

$\frac{d}{dx}\sin(x) \bigg|_{x=0}=1$

$\lim_{x \to 0}[f(x)]^2=\left( \lim_{x \to 0} f(x)\right)\left( \lim_{x \to 0} f(x)\right)=1\cdot 1=1$

Thats a really complex way to put it but i understand