# Find the limit. Use l'Hospital's Rule if appropriate. limit goes to zero. f(x)=(x)/(tan^-1(5x))

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### 1 Answer

`lim_(x->0) x/(tan^-1(5x))`

Since x->0 and tan^(-1)(5x)->0 as x->0 we can use L'Hosptital's rule

`(dx)/(dx)=1` and `(d)/(dx)tan^-1(5x) = 1/(1+(5x)^2)*(d(5x))/(dx) = 5/(25x^2+1)`

So `lim_(x->0)x/(tan^(-1)(5x)) = lim_(x->0) 1/((5/(25x^2+1))) = lim_(x->0) (25x^2+1)/5`

this last limit can be evaluated at x=0 and is = 1/5

So

`lim_(x->0)x/(tan^(-1)(5x)) =1/5`