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Right-hand limit of arctan(1/x) as x -> 0+

Consider the function f(x)=arctan⁡(1x)f(x) = \arctan\left(\dfrac{1}{x}\right), defined for all x≠0x \neq 0. We want to investigate its behavior as xx approaches 00.

As x→0+x \to 0^+, the quantity 1/x1/x grows without bound (approaches +∞+\infty). Using this fact, along with the graph of y=arctan⁡(u)y = \arctan(u) below, determine lim⁡x→0+arctan⁡(1x).\lim_{x \to 0^+} \arctan\left(\frac{1}{x}\right).

Right-hand limit of arctan(1/x) as x -> 0+ graph
Hint

Think about what happens to arctan⁡(u)\arctan(u) as uu becomes a very large positive number. Recall the horizontal asymptote of the arctangent function on the right side of its graph.

Step-by-step solution

As x→0+x \to 0^+, the value of xx is small and positive, so 1/x1/x becomes a large positive number, tending toward +∞+\infty. So we need to know how arctan⁡(u)\arctan(u) behaves as u→+∞u \to +\infty.

From the graph, arctan⁡(u)\arctan(u) approaches the horizontal asymptote y=π/2y = \pi/2 as u→+∞u \to +\infty, but never actually reaches it. Therefore, lim⁡x→0+arctan⁡(1/x)=π/2\displaystyle\lim_{x \to 0^+} \arctan(1/x) = \pi/2.

Left-hand limit of arctan(1/x) as x -> 0-

As x→0−x \to 0^-, the quantity 1/x1/x decreases without bound (approaches −∞-\infty). Using this fact, along with the graph of y=arctan⁡(u)y = \arctan(u) shown in part (a), determine lim⁡x→0−arctan⁡(1x).\lim_{x \to 0^-} \arctan\left(\frac{1}{x}\right).

Hint

Consider what happens to arctan⁡(u)\arctan(u) as uu becomes a very large negative number. Look at the other horizontal asymptote of the arctangent graph.

Step-by-step solution

As x→0−x \to 0^-, xx is small and negative, so 1/x1/x becomes a large negative number, tending toward −∞-\infty. We need the behavior of arctan⁡(u)\arctan(u) as u→−∞u \to -\infty.

From the graph, arctan⁡(u)\arctan(u) approaches the horizontal asymptote y=−π/2y = -\pi/2 as u→−∞u \to -\infty. Therefore, lim⁡x→0−arctan⁡(1/x)=−π/2\displaystyle\lim_{x \to 0^-} \arctan(1/x) = -\pi/2.

Existence of the two-sided limit

Based on your results from parts (a) and (b), does lim⁡x→0arctan⁡(1x)\displaystyle\lim_{x \to 0} \arctan\left(\frac{1}{x}\right) exist? Justify your answer briefly.

Hint

Recall the rule connecting one-sided limits to the two-sided limit: they must both exist and be equal.

Step-by-step solution

A two-sided limit exists if and only if the left-hand and right-hand limits both exist and are equal to each other.

Here the right-hand limit is π/2\pi/2 and the left-hand limit is −π/2-\pi/2, which are not equal. Therefore lim⁡x→0arctan⁡(1/x)\displaystyle\lim_{x \to 0} \arctan(1/x) does not exist.

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