Tutorials › Calculus I › One-Sided Limits At Radical Endpoints

One-Sided and Infinite Limits · Tutorial 235 of 1495

One-Sided Limits At Radical Endpoints

Determine which side of a radical endpoint is allowed, then use algebra and signs to evaluate finite or infinite one-sided limits.

Intermediate 10 min read

What You'll Learn

  • Identify the allowed side of an endpoint from the radicand
  • Distinguish a missing one-sided limit from an infinite one-sided limit
  • Track the sign of a denominator approaching zero
  • Use conjugates to simplify quotients with square roots
  • Recognize when a zero-over-zero form needs further analysis

Introduction

A square root restricts the inputs of a real-valued function: its radicand must be nonnegative. When the radicand becomes zero at an endpoint, the function may be defined on only one side of that endpoint. A one-sided limit must respect this domain restriction.

The domain tells us which direction to approach from. After that, the sign and size of the expression determine whether the limit is finite, infinite, or does not exist. In particular, a denominator approaching zero can produce either positive or negative infinity, depending on its sign on the allowed side.

Which Side of the Endpoint Is Allowed?

For a real square root \(\sqrt{g(x)}\), the radicand must satisfy \(g(x)\ge 0\). If \(g(x)=x-a\), this condition is \(x\ge a\), so the domain begins at \(a\) and points to the right. If \(g(x)=a-x\), the condition is \(x\le a\), so the domain ends at \(a\) and points to the left.

Definition: A right-hand limit \(\lim_{x\to a^+}f(x)\) describes the values of \(f(x)\) as \(x\) approaches \(a\) through inputs greater than \(a\). A left-hand limit \(\lim_{x\to a^-}f(x)\) uses inputs less than \(a\). At a domain endpoint, only the side contained in the domain can be used.

For example, \(f(x)=\sqrt{x-2}\) is real-valued only for \(x\ge 2\), so \(\lim_{x\to2^+}f(x)\) is meaningful, but there are no domain inputs immediately to the left of \(2\) on which to form \(\lim_{x\to2^-}f(x)\). The missing left-hand limit is not a limit equal to zero or infinity; the function simply has no values on that side near the endpoint.

When an affine radicand is written as \(c(x-a)\) with \(c>0\), its nonnegativity also requires \(x\ge a\). If it is \(c(a-x)\) with \(c>0\), it requires \(x\le a\). For more complicated radicands, determine the sign near the endpoint before choosing a side.

Roots Approach Zero, but Their Reciprocals Can Grow Without Bound

If \(t\ge0\) and \(t\to0\), then \(\sqrt{t}\to0\). This follows directly from the definition of a limit: for any \(\varepsilon>0\), if \(0\le t<\varepsilon^2\), then \(0\le\sqrt{t}<\varepsilon\). The sign restriction matters. A real square root near a zero radicand is nonnegative.

By contrast, when \(t\to0^+\), the reciprocal \(1/\sqrt{t}\) grows without bound. To see this, take any \(M>0\). If \(0<t<1/M^2\), then \(\sqrt{t}<1/M\), and hence \(1/\sqrt{t}>M\). Thus the reciprocal tends to \(+\infty\).

Theorem: For \(t>0\), \(\frac{\sqrt{t}}{t}=\frac{1}{\sqrt{t}}\). Consequently, as \(t\to0^+\), \(\frac{\sqrt{t}}{t}\to+\infty\). If the denominator is \(-t\) instead, the quotient is negative and tends to \(-\infty\).

This simple comparison is useful, but not every quotient at an endpoint has this form. If both numerator and denominator approach zero, the result cannot be read off from those two facts alone. Algebra, often using a conjugate, can reveal the behavior.

Worked Example: A Positive Infinite Limit at a Right Endpoint

Worked Example: A Positive Infinite Limit at a Right Endpoint

Evaluate \(\lim_{x\to0^+}\frac{\sqrt{x}}{x}\).

The radicand requires \(x\ge0\), and the denominator excludes \(x=0\). Thus the function is defined immediately to the right of \(0\), which is the allowed approach side. For \(x>0\), write \(x=(\sqrt{x})^2\), so

$$ \frac{\sqrt{x}}{x} = \frac{\sqrt{x}}{(\sqrt{x})^2} = \frac{1}{\sqrt{x}}. $$

As \(x\to0^+\), the positive quantity \(\sqrt{x}\) tends to zero, so its reciprocal tends to \(+\infty\). Therefore,

$$ \lim_{x\to0^+}\frac{\sqrt{x}}{x}=+\infty. $$

Check: For \(x=0.01\), the original quotient is \(0.1/0.01=10\), and the simplified quotient is \(1/0.1=10\). For \(x=0.0001\), both forms give \(100\). The identity is valid for every \(x>0\), and the values grow positively as \(x\) approaches zero from the allowed side.

Track the Denominator’s Sign

A denominator tending to zero does not determine the sign of an infinite limit by itself. Check whether it approaches zero through positive or negative values on the allowed side. The numerator matters too: if it stays positive, the quotient has the denominator’s sign; if it stays negative, the quotient has the opposite sign.

Worked Example: A Negative Infinite Limit at a Left Endpoint

Evaluate \(\lim_{x\to4^-}\frac{\sqrt{4-x}}{x-4}\).

The radicand condition \(4-x\ge0\) requires \(x\le4\), so the allowed side is the left. For \(x<4\), the numerator is positive and \(x-4=-(4-x)\). Therefore,

$$ \frac{\sqrt{4-x}}{x-4} = -\frac{\sqrt{4-x}}{4-x} = -\frac{1}{\sqrt{4-x}}. $$

As \(x\to4^-\), \(4-x\to0^+\), so \(1/\sqrt{4-x}\to+\infty\). The minus sign makes the original quotient tend to \(-\infty\):

$$ \lim_{x\to4^-}\frac{\sqrt{4-x}}{x-4}=-\infty. $$

Check: At \(x=3.99\), the numerator is \(\sqrt{0.01}=0.1\), the denominator is \(-0.01\), and the quotient is \(-10\). The simplified form gives \(-1/\sqrt{0.01}=-10\). The sign and the increasing magnitude agree with the limit.

The same sign check applies when a square root is only part of a more complicated denominator. Use the domain to determine which side is available, and then use the factors or a sign chart to establish whether the denominator is positive or negative there.

When Both Numerator and Denominator Approach Zero

If both parts of a quotient approach zero, direct substitution gives the indeterminate form \(0/0\). This does not specify the limit. For radical expressions, multiplying by the conjugate often exposes a factor that can be canceled. The conjugate identity works because \((u-v)(u+v)=u^2-v^2\).

Method: When a numerator contains a difference of square roots, multiply the numerator and denominator by the corresponding sum. Simplify using the difference-of-squares identity, and retain the original domain restrictions when canceling factors.

Worked Example: A Finite Limit After Rationalizing

Evaluate \(\lim_{x\to-1^+}\frac{\sqrt{x+2}-1}{\sqrt{x+1}}\).

The denominator requires \(x+1\ge0\), and the numerator’s radicand requires \(x+2\ge0\). Near \(-1\), the combined domain is \(x\ge-1\), so the allowed approach is from the right. Direct substitution gives a zero numerator and a zero denominator, so it does not settle the limit.

For \(x>-1\), multiply by the conjugate of the numerator:

$$ \frac{\sqrt{x+2}-1}{\sqrt{x+1}} \cdot \frac{\sqrt{x+2}+1}{\sqrt{x+2}+1} = \frac{(x+2)-1}{\sqrt{x+1}(\sqrt{x+2}+1)} = \frac{x+1}{\sqrt{x+1}(\sqrt{x+2}+1)}. $$

Since \(x>-1\), \(\sqrt{x+1}>0\), and \(x+1=(\sqrt{x+1})^2\). Cancel one positive factor to obtain

$$ \frac{\sqrt{x+1}}{\sqrt{x+2}+1}. $$

Now as \(x\to-1^+\), the numerator tends to \(0\) and the denominator tends to \(1+1=2\). Hence

$$ \lim_{x\to-1^+}\frac{\sqrt{x+2}-1}{\sqrt{x+1}}=0. $$

Check: At \(x=-0.99\), the original expression is \((\sqrt{1.01}-1)/\sqrt{0.01}\). Rationalizing gives \(\sqrt{0.01}/(\sqrt{1.01}+1)\), the same value because the rationalization identity holds for every \(x>-1\). As \(x\) approaches \(-1\) from above, this simplified expression tends to \(0/2=0\), confirming the finite limit.

A Practical Endpoint Procedure

For a one-sided limit involving a square root at a domain endpoint, keep the domain and the limit calculation separate at first. The domain decides which inputs can approach the endpoint; the algebra and signs decide what the function values do.

1
Find the domain near the endpoint.
Require every square-root radicand to be nonnegative and every denominator to be nonzero.
2
Identify the available side.
Determine whether nearby domain inputs lie to the left or right. Do not try to evaluate a one-sided limit from a side with no domain points.
3
Check numerator and denominator behavior.
Determine which tend to zero and, for a denominator near zero, whether it is positive or negative on the allowed side.
4
Simplify if necessary.
If direct substitution produces \(0/0\), try a conjugate or another valid algebraic simplification. Keep track of where any cancellation is permitted.
5
State the conclusion from the allowed side.
Distinguish a finite value, \(+\infty\), \(-\infty\), and the absence of an approach through the domain.

Common Mistake: Ignoring the Allowed Side’s Sign

In the left-endpoint example, a common error is to treat \(x-4\) as positive because it is a denominator close to zero. That gives an incorrect positive infinite limit. But for every \(x<4\), \(x-4<0\). The numerator \(\sqrt{4-x}\) is positive, so the quotient is negative.

$$ x<4 \quad\Longrightarrow\quad x-4<0, \qquad \sqrt{4-x}>0, \qquad \frac{\sqrt{4-x}}{x-4}<0. $$

The correction is to use \(x-4=-(4-x)\), which gives \(-1/\sqrt{4-x}\) and therefore the limit \(-\infty\). The sign is not a detail to check after the calculation: it is part of the limit analysis.

Why the Domain Restriction Matters

At a radical endpoint, domain information determines whether a one-sided limit is available at all. For example, \(\sqrt{x-a}\) has real values only for \(x\ge a\); a left-hand approach through the real domain is impossible. If a denominator also approaches zero, its sign on the permitted side determines whether an unbounded quotient tends to \(+\infty\) or \(-\infty\). If both numerator and denominator vanish, a conjugate can reveal a finite limit or a different behavior.

These checks prevent two distinct errors: claiming a limit from a side where the function is not defined, and assigning an infinite limit without verifying its sign. They also prepare you to analyze the behavior of functions near points where values grow without bound.

Check Your Understanding

For each question, identify the allowed approach side before evaluating or classifying the limit.

  1. What side of \(a\) is available for \(\sqrt{x-a}\) near \(x=a\)? What about \(\sqrt{a-x}\)?
  2. Evaluate \(\lim_{x\to0^+}\frac{\sqrt{x}}{x}\). Why is the limit positive?
  3. Evaluate \(\lim_{x\to9^-}\frac{\sqrt{9-x}}{x-9}\). What determines the sign?
  4. For \(\lim_{x\to-1^+}\frac{\sqrt{x+2}-1}{\sqrt{x+1}}\), why is direct substitution insufficient, and what algebraic identity helps?
  5. Does \(\sqrt{x-3}\) have a real-valued left-hand limit at \(x=3\)? Explain using its domain.

Test yourself: Calculus I final exam practice

Ready to check what you know? Work a full-length practice final, then open the complete worked solutions.

Want all 1,114 practice problems?

This tutorial covers one topic. Calc Pass covers the full Calculus AB & BC curriculum (the core of Calculus I and II): full review, every practice question, and step-by-step solutions to each one.

QR code to download the Calc Pass app

Get the Calc Pass App

Scan to download and practice on the go, with instant scoring.

Calc Pass Volume 1: 1,114 Practice Questions book cover

Vol. 1: 1,114 Questions

The complete practice question bank, organized by AB & BC topic.

View on Amazon
Calc Pass Volume 2: Solutions Manual book cover

Vol. 2: Solutions Manual

Fully worked solutions to every question in Volume 1.

View on Amazon