lim as x -->o+ (1/sin(x) - 1/x) ?

Question

Here's my thinking: As x-->0+, 1/sin(x)--> 1 and -1/x -->neg infinity. So is the lim "neg infinity"? Please explain and tell me if I am drawing an incorrect conclusion. Thank you.

Answer 1

as x-->0 sinx -->0 and 1/sin(x) -->oo
you can't just say +oo - oo = anything
you have to show that they both are diverging in synch.

sinx ~ x - x^3/6 +O(x^5)
so 1/sinx ~ (1/x) ( 1-x^2/6+O(x^4))^(-1)
~ (1/x)(1+x^2/6 +O(x^4))
= 1/x + x/6 + O(x^3)
so 1/sinx - 1/x ~ x/6 +O(x^3) --> 0

Answer 2

1/sin(x) - 1/x = (x - sin(x))/(x sin(x)). We know that sin(x) = x -x^3/3! + x^5/5! - x^7/7! .......

So, 1/sin(x) - 1/x = (x^3/3! - x^5/5! + x^7/7! ......)(x sin(x)) = (x^3/3! - x^5/5! + x^7/7! ......)(x^2 - x^4/3! + x^6/6!....) Since the leading term in the numerator has degree 3 and the leading term in the denominator has degree 2, it follows the ration goes to 0 as x -> 0. Therefore, lim (x -> 0) (1/sin(x) - 1/x) = 0

Your thinking is not correct because both 1/x and i/sin(x) goes to oo when -> 0+, so we have an undefined limit of the type oo - oo.

Answer 3

so far none has answered correctly

1/ sin x - 1/x = (x- sin x)/ x sinx

as x ->0 it is of the form zero by zero so

differentiating both num and denominator

(1- cos x)/( x cos x + sin x)
again it is 0/0
so continuing

sin x/( - x sin x + cos x + cos x)

numerator is zero and denominator is 2 so limit is 0

Answer 4

1/sinx does not go to 1 as x goes to zero.

As x goes to zero, sinx goes to x.

So the expression becomes 1/x - 1/x = 0.

It goes to 0 and x goes to 0.

You could verify this on your calculator by putting x = 0.001 or -0.001. Make sure you work in radians, not degrees.

Answer 5

by L'Hopital's rule
lim x->0 1/sin(x) = lim x->0 {(d1/dx) / (d sin(x)/dx) }= 0

lim x->0 1/x = lim x->0 {(d1/dx) / (dx/dx)} = 0

so lim x->0 (1/sin(x) - 1/x) = 0

maybe you confused lim x->0 {x/sin(x)} with lim x->0 {sin(x)/x}, the second one is 1

Answer 6

sin(0) = 0, not 1.

So, you have to use L'Hospital's rule.