Evaluate: lim_(x→0) (arcsinx-x)/x^3?

2 Answers
Feb 7, 2018

0.

Explanation:

We know that, (1): lim_(theta to 0)theta/sintheta=1.

(2):sin3theta=3sintheta-4sin^3theta.

Let, arc sinx=theta," so that, "x=sintheta.

Also, as x to 0, theta to 0, and sintheta to 0," as well".

:."The Reqd. Lim."=lim_(x to 0)(arc sinx-x)/x^3,

=lim_(theta to 0)(theta-sintheta)/sin^3theta,

=lim(theta-sintheta)/{1/4(3sintheta-sin3theta)}......[because, (2)],

=lim{4(theta-sintheta)}/(3sintheta-sin3theta),

=lim{4sintheta(theta/sintheta-1)}/{sintheta(3-sin^2theta)},

=lim_(theta to 0)(theta/sintheta-1)/(3-sin^2theta),

=(1-1)/(3-0^2)............[because, (1)].

rArr" The Reqd. Lim"=0.

Feb 7, 2018

1/6

Explanation:

Substituting 0 in the given function we get a 0/0 indeterminate form.
We can use the L'Hospital's Rule,
lim_(x→0) (1/sqrt(1-x^2)-1)/(3x^2)

lim_(x→0) ((1-sqrt(1-x^2))/sqrt(1-x^2))/(3x^2)

lim_(x→0) (1-sqrt(1-x^2))/(3x^2) . lim_(x -> 0) (1/sqrt(1-x^2))

Evaluating second limit which is 1 and applying L'Hospital's Rule to first limit,

lim_(x→0) (-(-2x)/(2sqrt(1-x^2)))/(6x)

lim_(x→0) 1/(6(sqrt(1-x^2))

Substituting 0 and we get 1/6