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

2 Answers
Jul 16, 2016

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

Explanation:

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]}

Jul 17, 2016

A lengthier approach using trig is explained below.

Explanation:

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:
enter image source here
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