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

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Answer 1 · 2018-05-15T22:26:19

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

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 · 2018-05-16T11:45:20

3

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:

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

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