Introduction

One-sided limits arise when we encounter limits that could only exist from the left or right, but not from both sides. The ones we have studied so far are precisely of this last kind: for the limit of the function f when x\to x_0 to exist, it is necessary that f be well-defined on both sides of x_0; if this does not happen, then the definition of the limit will not work. Since situations where such limits occur are frequent, it is necessary to find a way to deal with them. This is resolved through a formal definition.

Intuitive Idea of One-Sided and Bilateral Limits

For the limit of a function to exist f, when x\to x_0, it is necessary that the function be well-defined on both sides of x_0. If this occurs, then we talk about a bilateral limit. And if such a limit also results in L, then there would be no problem writing

\displaystyle \lim_{x\to x_0}f(x) = L

Now, imagine that we redefine this function in such a way that its domain only includes values greater than x_0. If we do this, we will notice that the limit has ceased to exist (because there will be values of x for which it will not make sense); however, graphically we could still say that when x\to x_0, f(x) still tends to L. The intuitive idea we are producing here is that of a right-hand limit, which is what we would represent through the notation

\displaystyle \lim_{x\to x_0^+}f(x) = L

and similarly, we will have the left-hand limit

\displaystyle \lim_{x\to x_0^-}f(x) = L

Finally, the bilateral limit will exist whenever the one-sided limits exist and are equal

\displaystyle \lim_{x\to x_0^-}f(x) = \lim_{x\to x_0}f(x) = \lim_{x\to x_0^-}f(x)

Formal Definition of One-Sided Limits

To formally define one-sided limits, it is sufficient to apply a small modification to the original limit definition.

\displaystyle \lim_{x\to x_0} f(x) = L := \left(\forall \epsilon\gt 0 \right) \left(\exist \delta \gt 0 \right) \left(\right. 0\lt|x-x_0|\lt\delta \left.\rightarrow |f(x)-L|\lt\epsilon \right)

For right-hand limits, the definition is as follows:

\left(\forall \epsilon\gt 0 \right) \left(\exist \delta \gt 0 \right) \left(\right. x_0 \lt x \lt x_0 + \delta \left.\rightarrow |f(x)-L|\lt\epsilon \right)

For left-hand limits, it will be:

\left(\forall \epsilon\gt 0 \right) \left(\exist \delta \gt 0 \right) \left(\right. x_0 - \delta \lt x \lt x_0 \left.\rightarrow |f(x)-L|\lt\epsilon \right)

Conditions for Limit Algebra

The interesting thing about having these definitions is that they are both contained within the usual limit definition, and this is important because it frees us from having to re-prove all the properties we already proved for bilateral limits. All the limit algebra will work just as we have seen in previous classes, as long as the limits involved are of the same nature (both from the left, or both from the right, never mixed), are directed to the same point, and exist at that point.

Proposed and Solved Exercises

1. \displaystyle \lim_{x\to {\frac{1}{2}}^- } \sqrt{\dfrac{x+2}{x+1}} [SOLUTION]
2. \displaystyle \lim_{x\to 1^+} \sqrt{\dfrac{x-1}{x+2}} [SOLUTION]
3. \displaystyle \lim_{x\to 2^+} \left(\dfrac{x}{x+1} \right) \left(\dfrac{2x+5}{x^2+x} \right) [SOLUTION]
4. \displaystyle \lim_{x\to 1^-} \left(\dfrac{1}{x+1} \right) \left(\dfrac{x+6}{x} \right) \left(\dfrac{3-x}{x} \right) [SOLUTION]
5. \displaystyle \lim_{h\to 0^+ } \dfrac{\sqrt{h^2 + 4h +5} - \sqrt{5}}{h} [SOLUTION]
6. \displaystyle \lim_{h\to 0^-} \dfrac{\sqrt{6} - \sqrt{5h^2 + 11h +6}}{h} [SOLUTION]

7. a. \displaystyle \lim_{x\to -2^+} (x+3)\dfrac{|x+2|}{x+2}

   b. \displaystyle \lim_{x\to -2^-} (x+3)\dfrac{|x+2|}{x+2}

   [SOLUTION]

8. a. \displaystyle \lim_{x\to 1^+} \dfrac{\sqrt{2x}(x-1)}{|x-1|}

   b. \displaystyle \lim_{x\to 1^-}\dfrac{\sqrt{2x}(x-1)}{|x-1|}

   [SOLUTION]