Question

lim x tends to infinity (1/sinx-1/x)

Answer 1

1x−1sinx=sinx−xxsinx

Now, Taylor’s theorem tells us that sinx in the vicinity of zero is:

sin(x)=x−x36+h(x)x3

for some function h such that limx→0+h(x)=0.

Using this result we get:

limx→0+1x−1sinx=limx→0+sinx−xxsinx

=limx→0+x−x36+h(x)x3−xx(x−x36+h(x)x3)

=limx→0+−x36+h(x)x3x2−x46+h(x)x4

=limx→0+−x6+h(x)x1−x26+h(x)x2

Now the denominator clearly limits to 1 and the numerator limits to zero, which implies that the final answer must be zero.