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

**petike** · Oct 19th 2008, 05:00 AM

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

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

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

Thanks.

**Plato** · Oct 19th 2008, 05:05 AM

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

**petike** · Oct 19th 2008, 06:03 AM

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

**ThePerfectHacker** · Oct 19th 2008, 06:37 AM

Originally Posted by petike

"indeterminate expression", because it is equal to: $\frac{0}{0}$

No it is not equal to that.

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

**Plato** · Oct 19th 2008, 08:28 AM

Originally Posted by petike

I think that $\lim_{x \rightarrow 0} \frac{x}{x} = \mbox{"Indeterminate/expression"}$ , 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 $\lim _{x \to x_0 } f(x)$ it is the case that $x$ never, never, never equals $x_0$ !
So in the limit $\lim _{x \to 0 } \frac{x} {x}$ it is true that $x$ never, never, never equals 0.
So if $x \ne 0\$ then $\frac{x} {x} = 1$ so that is the limit.

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

**rman144** · Oct 19th 2008, 08:42 AM

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

So your solution is:

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

Since cos(0)=1, you have:

1/1=1

**ThePerfectHacker** · Oct 19th 2008, 08:45 AM

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.

**rman144** · Oct 19th 2008, 08:50 AM

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.

**petike** · Oct 19th 2008, 11:59 PM

Thank you for all your answers.
The best help I have maybe given from "ThePerfectHacker":

Originally Posted by ThePerfectHacker

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

Once more, thanks.

**qorilla** · Oct 20th 2008, 07:03 AM

What your teacher meant with "indeterminate expression" is the following.
If you only know about a fraction, that the numerator and the denominator both approach 0, then the limit of the quotient could be anything (+/- infinity or 0 or 32 or 1347.312 or any other real number)
Knowing that it's in the form 0/0 is no information about it's limit.
It is indeterminate in a sense that the 0/0 form does not determine the limit of the fraction like for example 2/3 does (if the numerator approaches 2 and the denominator approaches 3, the whole fraction approaches 2/3 for sure).

**skamoni** · Oct 21st 2008, 03:58 PM

sinx/x is bounded by 1/cosx and cosx. So apply what limit techniques you know to these relatively simple functions, and then by the "sandwhich" rule, you have the answer.

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