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# `lim_(x -> 0) (sin^3(x-pi))/x^3` is a. -1 b.  0 c. 1 d. `pi` e. Non-exisitent

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`lim_(x -> 0) (sin^3(x-pi))/x^3` is

a. -1

b.  0

c. 1

d. `pi`

e. Non-exisitent

Posted by draw on August 18, 2013 at 9:08 PM via web and tagged with calculus, math

Student

(Level 1) Associate Educator, Expert

(Level 1) Associate Educator

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`lim_(x->0) (sin^3(x-pi))/(x^3)` is of indeterminate type 0/0 when you plug-in 0 to x. Hence, we use L'Hospital's Rule:

`lim_(x->0) (sin^3(x-pi))/x^3 = lim_(x->0) ((d/(dx))sin^3(x-pi))/(d/(dx) x^3) = lim_(x->0) -(cosx*sin^2x)/x^2`

The resulting expression, when 0 is again plugged into x, is another indeterminate type 0/0. We use L'Hospital's Rule again:

`lim_(x->0) -(cosx*sin^2x)/x^2 = - (d/(dx) cosx*sin^2x)/(d/(dx) x^2) = lim_(x->0) (sinx - 3sin(3x))/(8x) `

`= 1/8 lim_(x->0) (sinx - 3sin(3x))/x`

Again, this resulting expression is indeterminate type 0/0. Hence, we shall apply L'Hospital's rule for the 3rd time.

`1/8 lim_(x->0) (sinx - 3sin(3x))/x = 1/8 lim_(x->0) (d/(dx) sinx - 3sin(3x))/(d/(dx) x) `

`= 1/8 lim_(x->0) (cosx - 9cos(3x))`

Since the limit of a sum is just the sum of the limits, we can re-write this last expression as:

`1/8 [lim_(x->0) cos x - lim_(x->0) 9cos(3x)]`

`1/8 [ lim_(x->0) cosx - 9 lim_(x->0) cos(3x)]`

`lim_(x->0)cosx = cos(0) = 1`

Hence,

`1/8 [ lim_(x->0) cosx - 9 lim_(x->0) cos(3x)] = 1/8[ 1- 9lim_(x->0)cos(3x)]`

Since the function f(x) = cos(x) is continuous at x = 0,

`lim_(x->0) cos(3x) = cos(lim_(x->0) 3x) = cos(3 lim_(x->0) x)=cos(0) = 1`

Hence,

`1/8 [ lim_(x->0) cosx - 9 lim_(x->0) cos(3x)] = 1/8 [1 - 9(1)]`

`= 1/8(-8) = -1`

Therefore:

`lim_(x->0) (sin^3(x-pi))/(x^3) = -1`

The limit of the given expression is -1.

Notes:

1. L'Hospital's Rule can only be used when the indeterminate form is infinity/infinity or 0/0.
2. If f(x) is continuous at x = a, you can interchange the limit and the function evaluation.
3. Evaluating limit of sums is equivalent to evaluating sum of limits.

Posted by mvcdc on August 18, 2013 at 9:45 PM (Answer #1)

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Find `lim_(x->0)(sin^3(x-pi))/x^3` :

`sin(x-pi)=-sinx` ; substituting we get:

`lim_(x->0)(sin^3(x-pi))/x^3=lim_(x->0)(-sin^3x)/x^3`

``Since `lim_(x->0)(sinx)/x=1` we get `lim_(x->0)[(-sinx)/x * (-sinx)/x * (-sinx)/x]=-1`