How do you find the limit of #(arctan(x)) / (5x)# as x approaches 0?

Question

Ezekiel Alexander · Accepted Answer

#lim_(x->0) (arctan x)/(5x) = 1/5#

To find this limit, notice that both the numerator and denominator go to #0# as #x# approaches #0#. This means that we'd get an indeterminate form, thus we can apply L'Hospital's rule.

#lim_(x->0) (arctan x)/(5x) -> 0/0#

By applying L'Hospital's rule, we take the derivative of the numerator and denominator, giving us

#lim_(x->0) (1/(x^2+1))/(5) = lim_(x->0) 1/(5x^2+5) = 1/(5(0)^2+5) = 1/5#

We can also check this by graphing the function, to get an idea what #x# approaches.

Graph of #arctan x / (5x)#: graph{(arctan x)/(5x) [-0.4536, 0.482, -0.0653, 0.4025]}

Elijah Abram · Answer

A lengthier approach using trig is explained below.

Just in case you're not comfortable with L'Hopital's Rule, or have not yet been exposed to it, another approach to solving the problem involves using the definition of the arctangent function.

Recall that if #tantheta=x#, then #theta=arctanx#; this essentially means that arctangent is the reverse of tangent. Using this info, we can construct a triangle where #tantheta=x# and #theta=arctanx#:

Axel Alvarez · Answer

undefined To find the limit of (arctan(x)) / (5x) as x approaches 0, we can use L'Hôpital's rule. Taking the derivative of both the numerator and denominator, we get 1 / (1 + x^2) / 5. Substituting x = 0 into this expression, we find that the limit is 1 / 5.