Explain: As x Approaches 0, Sinx/x & Tanx/x

[Wiz]

tends to 1 from??

i am confused on this one...my friend told me that as x->0 ,(sinx)/x tends to 1 from left hand side...similarly he said something abt tanx/x approaching 1 from right hand side...can anyone please explain this to me??
thanks..
wiz

 

[TD]

$\frac{{\sin x}}{x}$ tends to 1 when x goes to 0 from either left or right, therefore the limit exists and is 1.

Same goes for $\frac{{\tan x}}{x}$, it's valid for both sides so the limit is 1.

Btw, for small x, $\tan \left( x \right) \approx \sin \left( x \right)$

 

[AKG]

A function f approaches a limit L as x approaches a value b from the right hand side if, for every positive number E, you can find a positive number d such that for all x in (a, a + d), |f(x) - L| < E. The same is true for the left hand side, but then you'd want to find a positive number d such that for all x in (a - d, a), |f(x) - L| < E. In simpler terms, numbers on the right hand side are greater than a certain number, which makes sense (on the number line, numbers increase to the right and decrease to the left). Taking a simpler example, try f(x) = 2x. This function approaches 12 as x approaches 6 from the left-hand side (it also approaches 12 as x approaches 6 from the right-hand side, so we simply say that f(x) approaches 12 as x approaches 6). So if you take values for x of 5, 5.5, 5.9, 5.99, 5.999, 5.999999, etc., then your values for f(x) will get closer and closer to 12.

 

[lurflurf]

This is a common limit.
in fact it arises in finding the derivative of sine
$$\frac{d}{dx}\sin(x)=\lim_{h\rightarrow 0} \frac{\sin(x)}{x}\cos(x+\frac{h}{2})$$
the limit can be shown several ways including showing that
cos(x)<sin(x)/x<1 for |x|<pi/2
or seeing that for the function f
$$f(x):=\int_0^1 \cos(x t) dt$$
f is every where continuos and
f(x)=sin(x)/x for all x except x=0
(also xf(x)=sin(x))