find the limit as x approaches 0 (arcsin x)/x
To find the limit as x approaches 0 of (arcsin x)/x, we can use L'Hôpital's rule or the Taylor series expansion of arcsin x.
Using L'Hôpital's rule:
First, we can rewrite (arcsin x)/x as (1/x) * (arcsin x). Then, we can differentiate both the numerator and denominator with respect to x.
The derivative of arcsin x is 1/sqrt(1 - x^2), and the derivative of x is 1.
So, as x approaches 0, the limit becomes:
lim(x->0) (1/sqrt(1 - x^2)) * (1)
Now, we can substitute x = 0 into this expression, which gives us:
(1/sqrt(1 - 0^2)) * 1 = 1/sqrt(1) = 1/1 = 1.
Therefore, the limit as x approaches 0 of (arcsin x)/x is equal to 1.
Using the Taylor series expansion of arcsin x:
The Taylor series expansion of arcsin x is given by:
arcsin x = x + (x^3)/6 + (3x^5)/40 + ...
Substituting this expansion into (arcsin x)/x, we get:
(arcsin x)/x = (x + (x^3)/6 + (3x^5)/40 + ...) / x
= 1 + (x^2)/6 + (3x^4)/40 + ...
As x approaches 0, all the higher order terms (x^n where n > 1) become negligible since they approach 0 faster than the lower order terms. Therefore, the limit as x approaches 0 of (arcsin x)/x is equal to 1.
So, using either L'Hôpital's rule or the Taylor series expansion, we find that the limit as x approaches 0 of (arcsin x)/x is equal to 1.