# Calc. Find the limit using L'Hospital's Rule. lim x--->0 (x)/(tan^-1 (4x))

## Expert Answers

You need to use l'Hospital's theorem if the limit proves to be indeterminate, hence you should check if it is the case such that:

`lim_(x-gt0) x/(tan^(-1)4x) = 0/(tan^(-1) 0) = 0/0`

Since the limit is indeterminate, you may use l'Hospital's theorem such that:

`lim_(x-gt0) x/(tan^(-1)4x) = lim_(x-gt0) (x')/((tan^(-1)4x)')`

`lim_(x-gt0) (x')/((tan^(-1)4x)')...

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You need to use l'Hospital's theorem if the limit proves to be indeterminate, hence you should check if it is the case such that:

`lim_(x-gt0) x/(tan^(-1)4x) = 0/(tan^(-1) 0) = 0/0`

Since the limit is indeterminate, you may use l'Hospital's theorem such that:

`lim_(x-gt0) x/(tan^(-1)4x) = lim_(x-gt0) (x')/((tan^(-1)4x)')`

`lim_(x-gt0) (x')/((tan^(-1)4x)') = lim_(x-gt0) 1/(((4x)')/(1+16x^2))`

`lim_(x-gt0) 1/(((4x)')/(1+16x^2))= lim_(x-gt0)(1+16x^2)/4`

`lim_(x-gt0)(1+16x^2)/4 = (1+0)/4 = 1/4`

Hence, evaluating the limit to the given function yields `lim_(x-gt0) x/(tan^(-1)4x) =1/4.`

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