# Thread: "Limit of (sin x) over (x) = 1"?

1. ## "Limit of (sin x) over (x) = 1"?

Hi,
I have a little question. It is fact that:
$\displaystyle \lim_{x \rightarrow 0} \frac{sin x}{x} = 1$

But when I have a limit:
$\displaystyle \lim_{x \rightarrow 0} \frac{x}{x}$
it is considered an "indeterminate expression", because it is equal to: $\displaystyle \frac{0}{0}$
but I think that: $\displaystyle \lim_{x \rightarrow 0} \frac{sin x}{x} = \frac{0}{0}$ and it would be also an indeterminate expression.

So can anybody explain me why "is the first expession = 1"?

Thanks.

2. Do you understand that the following is equal to 1?
$\displaystyle \lim_{x \rightarrow 0} \frac{x}{x}={\color{blue}1}$

3. ## No...

No,
I think that
$\displaystyle \lim_{x \rightarrow 0} \frac{x}{x} = "Indeterminate \ expression"$, isn't it?

4. Originally Posted by petike
"indeterminate expression", because it is equal to: $\displaystyle \frac{0}{0}$
No it is not equal to that.

That is what asking what is the limiting value when $\displaystyle x$ is almost $\displaystyle 0$.
Take $\displaystyle x=.1,.01,.001,.0001,...$ and you will realize that $\displaystyle \tfrac{\sin x}{x}$ gets closer to $\displaystyle 1$.

5. Originally Posted by petike
I think that $\displaystyle \lim_{x \rightarrow 0} \frac{x}{x} = \mbox{"Indeterminate/expression"}$, isn't it?
It is certainly not indeterminate. That limit equals 1.
Some instructor has given you a great deal of misinformation.
Did you know that when evaluating limits such as $\displaystyle \lim _{x \to x_0 } f(x)$ it is the case that $\displaystyle x$ never, never, never equals $\displaystyle x_0$!
So in the limit $\displaystyle \lim _{x \to 0 } \frac{x} {x}$ it is true that $\displaystyle x$ never, never, never equals 0.
So if $\displaystyle x \ne 0\$ then $\displaystyle \frac{x} {x} = 1$ so that is the limit.

In the other limit $\displaystyle \lim _{x \to 0 } \frac{\sin (x)} {x}$ for any $\displaystyle x \approx _ \ne 0$ (close to zero but not equal to zero) it is true that $\displaystyle \sin (x) \approx x \ne 0$ so it follows that $\displaystyle \frac{{\sin (x)}}{x} \approx 1$.

Look up L'Hopital's Rule:

If you have an intermediate situation like sin(x)/x, lim x appr. 0, take the derivative of the numerator and the denominator independently:

d/dx(sin(x)) = cos(x)

d/dx(x) = 1

cos(x)/1, lim x appr. 0

Since cos(0)=1, you have:

1/1=1

7. Originally Posted by rman144
Look up L'Hopital's Rule:
The original poster is probably just starting to learn Calculus.
He will not know what L'Hopitals rule is.
And besides L'Hopital is a greatly abused rule - no need to apply it here.

True, most students will use it for proofs and not understand the circular logic, however, after spending hours doing the alternative methods, its not shocking how many see its attraction. Either way, your answer is 1.

9. ## Soluted

Take $\displaystyle x=.1,.01,.001,.0001,...$ and you will realize that $\displaystyle \tfrac{\sin x}{x}$ gets closer to $\displaystyle 1$.