Question

Can you solve this limit simply by l'hopitle rule or is there a more complex method by which you can go about this?

Answer 1

`lim_(x to 0) (x^2sin(x))/(sin(x)-xcos(x)) = 3`

Explanation:

Given: `lim_(x to 0) (x^2sin(x))/(sin(x)-xcos(x))`

The expression evaluated at 0 yields the indeterminate form `0/0`, therefore, the use of L'Hôpital's rule is warranted.

Apply L'Hôpital's rule by differentiating the numerator and the denominator:

`lim_(x to 0) ((d(x^2sin(x)))/dx)/((d(sin(x)-xcos(x)))/dx)`

`lim_(x to 0) (2xsin(x)+x^2cos(x))/(xsin(x))`

Simplify:

`lim_(x to 0) 2+(xcos(x))/sin(x)`

substitute `(xcos(x))/sin(x) = cos(x)/(sin(x)/x)`

`lim_(x to 0) 2+cos(x)/(sin(x)/x)`

`lim_(x to 0) sin(x)/x = 1` and `lim_(x to 0) cos(x) =1`, therefore, the fraction becomes `1/1`:

`lim_(x to 0) 2+cos(x)/(sin(x)/x)= 3`

According to L'Hôpital's rule, the original expression must go to the same limit.

Answer 2

3

Explanation:

Using series expansions.

`lim_(x to 0) (x^2sin(x))/(sin(x)-xcos(x))`

`= lim_(x to 0) (x^2(x - x^3/(3!) + bb O(x^5) ))/((x - x^3/(3!) + bb O(x^5))-x(1- x^2/(2!) + bbO(x^4))`

Simplifying a bit:

`= lim_(x to 0) ( x^3 + bb O(x^5) )/( - x^3/(3!) + x^3/(2!) + bbO(x^4))`

`= lim_(x to 0) ( 1 + bb O(x^2) )/( 1/3 + bbO(x)) = 3`