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

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]}

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#:

From the diagram, it is clear that #tantheta=x/1=x#. Since #tantheta=sintheta/costheta#, we can express this as:
#tantheta=x#
#->sintheta/costheta=x#

Using this plus the fact that #theta=arctanx#, we can make replacements in the limit:
#lim_(x->0) arctanx/(5x)#
#->lim_(theta->arctan0) theta/(5sintheta/costheta)#
#->lim_(theta->0) theta/(5sintheta/costheta)#

This is equivalent to:
#lim_(theta->0)1/5*lim_(theta->0)theta*lim_(theta->0)costheta/sintheta#

#->1/5*lim_(theta->0)theta/sintheta*lim_(theta->0)costheta#

We know that #lim_(x->0) sintheta/theta=1#; so #lim_(x->0)1/(sintheta/theta)=1/1# or #lim_(x->0)theta/sintheta=1#. And since #cos0=1#, the limit evaluates to:
#1/5*lim_(theta->0)theta/sintheta*lim_(theta->0)costheta#

#->1/5*(1)*(1)=1/5#