calculus...?

Question

lim sin(h) / h=1
as approaches to 0.

can anyone give me the solution why its limit is equal to 1.

Answer 1

Use the Taylor expansion for sin(h)
sin(h) = h - (h^3)/3! + (h^5)/5! - (h^7)/7! + ...

Divide by h
sin(h) / h = 1 - (h^2)/3! + (h^4)/5! - (h^6)/7! + ...
Apply the limit h approaches 0
[sin(h) / h] will approach 1.

Alternatively:
Use l'Hôpital's rule
sin(h)/h = 0/0 as h approaches 0
Differentiate the numerator and the denominator to get
cos(h)/1
cos(0) = 1
The limit is 1

Answer 2

Because sin(0)=0 and h/h=1

Simply put, although 0 cannot be in the denominator (cannot be divided), the same number divided by itself should equal to 1. But since 0 is not divisible by 0 but still approaches 1 because of the h/h=1 property, the solution to sin(0)/0 will get infinitely close to 1 but never really equaling to 1. Without a cutoff point, people can only say that its limit is equal to 1.

Hope that helped! :)

Answer 3

Another solution is to use L'Hopital's Rule, which states that for a function that evaluates to 0/0 at a given point (as sin(h)/h does as it approaches h=0) the limit can be determined by evaulating the derivatives of the top and bottom portions and dividing THEM instead. The derivative of sin(h) is cos(h), whose value is 1 at h=0, and the derivative of h is 1. Cos(h)/1 evaluates to 1 when h=0.

Answer 4

In this case (when we get 0/0 if we directly substitute), we can use l'Hôpital's rule (also spelled l'Hospital). We take the derivative of the top alone and the derivative of the bottom alone so we'll have: lim cos(h)/1 as h approaches 0 which is equal to 1/1=1 (when we put h=0)

Answer 5

the small angle formula says that for small h sin(h) = h

sin(h) = h -h^3/3!+h^5/5!-h^7/7!+...

for small h we can say h^3/3! ~ 0 and the same for all terms after it. So we can say sin(h)~h

And therefore dividing both sides by h we get sin(h)/h ~ 1 which becomes sin(h)/h = 1 as h approaches 0

Answer 6

the only acceptable answer is the one of Katsaonisvagel as application of L´Hôpital or Taylor needs the knowledge of the derivative of sin h at h = 0 which is just the limit you ate trying to find out.

Answer 7

For 0 sinx
1
cosx For -pi/2 cos(-x)
cosx Finally near 0 happens
cosx Applying Santwitch theorem
we take limsinx/x=1,x-->0