# Use lhopital's rule to evaluate the limit  `lim_(x-gt0)x^3/(x-tan(x))`

thilina-g | College Teacher | (Level 1) Educator

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`lim_(x-gt0)x^3/(x-tan(x))`

If you try to evaluate the limit straight awy you would get 0/0 which is indeterminate. In such situations, L'hopital's Rule is so useful.

It says, for functions `f(x)` and `g(x)` ,

`lim_(x->0)(f(x))/(g(x)) = lim_(x-a)(f'(x))/(g'(x))`

Provided that `g'(x)` is non zero at `x = a.`

`f'(x) = 3x^2`

`g'(x) = 1 - sec^(x)`

`g'(0)` is zero, we can't apply L'hopital's Rule here, but we can extend it as below.

`lim_(x->0)(f'(x))/(g'(x)) = lim_(x->0)(f''(x))/(g''(x))`

`f''(x) =6x`

`g''(x) = 0 - 2tan(x)sec^2(x) = - 2tan(x)sec^2(x)`

Still `g''(0) = 0.`

Therefore we have to extend the rule further more.

`lim_(x->0)(f''(x))/(g''(x)) = lim_(x->0)(f'''(x))/(g'''(x))`

`f'''(x) = 6`

`g'''(x) = -2(sec^2(x)sec^2(x)+tan(x)(- 2tan(x)sec^2(x))`

`g'''(x) = -2(sec^4(x)- 2tan^2(x)sec^2(x))`

`g'''(0) = -2(1-0) = -2`

Therefore, we can L'hopital's rule here,

Therefore,

`lim_(x->0)(f(x))/(g(x)) = lim_(x->0)(f'''(x))/(g'''(x))`

`lim_(x->0)(f(x))/(g(x)) = lim_(x->0)[6/(-2(sec^4(x)- 2tan^2(x)sec^2(x)))]`

`lim_(x->0)(f(x))/(g(x)) =[6/(-2(1-0))] = -3`

Therefore, `lim_(x-gt0)x^3/(x-tan(x)) = -3`

The limit is -3.