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Special Techniques for Limits at Infinity

Reference: Stewart 3.4 • Chapter: 3 • Section: 4

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Beyond Polynomial Ratios

Not every limit at infinity involves just polynomials. What happens when you have square roots, differences of large terms, or trigonometric functions? The "divide by highest power" technique needs adaptation, and sometimes entirely new approaches are required.

This skill covers three essential techniques:

Conjugate multiplication for radical expressions
Handling $\sqrt{x^2}$ when $x$ can be negative
Substitution for composition with trigonometric functions

These appear frequently on exams, often combined with rational function techniques.

Prerequisite Map

Quick Reference

Property	Value
Concept	Limits
Course	MATH161
Section	Stewart 2.6
Difficulty	Advanced
Time	~30 minutes

Key Concepts

Technique 1: Conjugate Multiplication

When to use: The expression involves a difference like $\sqrt{\text{something}} - \text{something}$, giving an indeterminate form "$\infty - \infty$".

The Method: Multiply numerator and denominator by the conjugate: the same expression with the sign switched.

Worked Example: Conjugate

Evaluate $\lim_{x \to \infty} \left(\sqrt{x^2 + 1} - x\right)$

Problem: Both $\sqrt{x^2 + 1}$ and $x$ go to infinity, so we get "$\infty - \infty$".

Step 1: Multiply by the conjugate over itself: $$\sqrt{x^2 + 1} - x = \frac{(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)}{\sqrt{x^2 + 1} + x}$$

Step 2: Use the difference of squares: $(a - b)(a + b) = a^2 - b^2$ $$= \frac{(x^2 + 1) - x^2}{\sqrt{x^2 + 1} + x} = \frac{1}{\sqrt{x^2 + 1} + x}$$

Step 3: Take the limit. As $x \to \infty$:

$\sqrt{x^2 + 1} \approx x$ (for large $x$)
Denominator $\approx x + x = 2x \to \infty$

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

Technique 2: Handling $\sqrt{x^2}$ for Negative $x$

Critical fact: $$\sqrt{x^2} = \vert x\vert = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$

When $x \to -\infty$: Since $x < 0$, we have $\sqrt{x^2} = -x$ (not $x$!).

This is the most common source of sign errors in limits at infinity.

Worked Example: Negative Infinity with Radicals

Evaluate $\lim_{x \to -\infty} \frac{\sqrt{2x^2 + 1}}{3x - 5}$

Step 1: Identify the highest power in the denominator: $x$.

Step 2: Divide numerator and denominator by $x$.

But wait: When $x < 0$, dividing by $x$ inside a square root requires care:

$$\frac{\sqrt{2x^2 + 1}}{x} = \frac{\sqrt{2x^2 + 1}}{\pm\sqrt{x^2}}$$

Since $x < 0$: $x = -\vert x\vert = -\sqrt{x^2}$

So: $$\frac{\sqrt{2x^2 + 1}}{x} = \frac{\sqrt{2x^2 + 1}}{-\sqrt{x^2}} = -\sqrt{\frac{2x^2 + 1}{x^2}} = -\sqrt{2 + \frac{1}{x^2}}$$

Step 3: Full expression: $$\frac{\sqrt{2x^2 + 1}}{3x - 5} = \frac{-\sqrt{2 + \frac{1}{x^2}}}{3 - \frac{5}{x}}$$

Step 4: Take limit as $x \to -\infty$: $$\lim_{x \to -\infty} \frac{-\sqrt{2 + \frac{1}{x^2}}}{3 - \frac{5}{x}} = \frac{-\sqrt{2 + 0}}{3 - 0} = -\frac{\sqrt{2}}{3}$$

Compare: For the same function as $x \to +\infty$: $$\lim_{x \to +\infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} = \frac{+\sqrt{2}}{3}$$

Different horizontal asymptotes on each side!

Visual Summary: $\sqrt{x^2}$ Sign Issue

                     √(x²) = |x|

    x < 0                           x > 0
    |x| = -x                        |x| = x
     ↙                               ↘

When dividing by x:            When dividing by x:
√(x²)/x = -x/x = -1           √(x²)/x = x/x = 1

The extra negative sign appears for x → -∞!

Technique 3: Substitution for Compositions

When to use: The function involves a composition like $\sin\left(\frac{1}{x}\right)$ or $e^{-x}$.

The Method: Substitute $t = \frac{1}{x}$ (or another appropriate variable) so that $x \to \infty$ becomes $t \to 0^+$.

Worked Example: Trigonometric Substitution

Evaluate $\lim_{x \to \infty} \sin\left(\frac{1}{x}\right)$

Step 1: Let $t = \frac{1}{x}$. As $x \to \infty$, we have $t \to 0^+$.

Step 2: Rewrite: $$\lim_{x \to \infty} \sin\left(\frac{1}{x}\right) = \lim_{t \to 0^+} \sin(t)$$

Step 3: Evaluate: $$\lim_{t \to 0^+} \sin(t) = \sin(0) = 0$$

When Limits Don't Exist

Oscillation without decay: If a function oscillates with constant amplitude as $x \to \infty$, the limit does not exist.

Example: $\lim_{x \to \infty} \sin(x)$ does not exist.

As $x$ increases, $\sin(x)$ keeps cycling between $-1$ and $1$ forever. It never settles to any single value.

Example: $\lim_{x \to \infty} \cos(x)$ does not exist (same reason).

Contrast: $\lim_{x \to \infty} \frac{\sin(x)}{x} = 0$ DOES exist.

Why? Even though $\sin(x)$ oscillates, it's bounded between $-1$ and $1$, while $x \to \infty$. So: $$-\frac{1}{x} \leq \frac{\sin(x)}{x} \leq \frac{1}{x}$$

By the Squeeze Theorem, the limit is $0$.

Practice Problems

Level 1 Recognizing Which Technique to Use

For each limit, identify which technique is most appropriate (don't solve):

$\lim_{x \to \infty} \left(\sqrt{x^2 + 5x} - x\right)$
$\lim_{x \to -\infty} \frac{\sqrt{9x^2 - 1}}{x + 4}$
$\lim_{x \to \infty} \cos\left(\frac{2}{x}\right)$
$\lim_{x \to \infty} (2x - 3)$

Thought Process

Look for:

Differences involving radicals → conjugate
Radicals with $x \to -\infty$ → careful with $\sqrt{x^2} = \vert x\vert = -x$
Compositions with trig → substitution
Simple polynomials → just evaluate directly

Show Answer

(a) $\sqrt{x^2 + 5x} - x$: Conjugate multiplication (Form is $\sqrt{\text{something}} - \text{something}$, giving $\infty - \infty$)

(b) $\frac{\sqrt{9x^2 - 1}}{x + 4}$ with $x \to -\infty$: $\sqrt{x^2} = -x$ technique (Radical with negative $x$ requires sign adjustment)

(c) $\cos\left(\frac{2}{x}\right)$: Substitution ($t = \frac{1}{x}$) (Composition with trig function)

(d) $2x - 3$: Direct evaluation (no special technique needed) This is just a polynomial: $\lim_{x \to \infty} (2x - 3) = \infty$

Level 2 Conjugate Multiplication

Evaluate $\lim_{x \to \infty} \left(\sqrt{x^2 + 3x} - x\right)$

Thought Process

This is "$\infty - \infty$". Multiply by the conjugate $\frac{\sqrt{x^2 + 3x} + x}{\sqrt{x^2 + 3x} + x}$ to clear the radical.

Show Answer

Multiply by the conjugate:

$$\sqrt{x^2 + 3x} - x = \frac{(\sqrt{x^2 + 3x} - x)(\sqrt{x^2 + 3x} + x)}{\sqrt{x^2 + 3x} + x}$$

Numerator becomes: $$(\sqrt{x^2 + 3x})^2 - x^2 = x^2 + 3x - x^2 = 3x$$

So: $$= \frac{3x}{\sqrt{x^2 + 3x} + x}$$

Now divide numerator and denominator by $x$ (with $x > 0$): $$= \frac{3}{\frac{\sqrt{x^2 + 3x}}{x} + 1} = \frac{3}{\sqrt{1 + \frac{3}{x}} + 1}$$

As $x \to \infty$: $$\lim_{x \to \infty} \frac{3}{\sqrt{1 + \frac{3}{x}} + 1} = \frac{3}{\sqrt{1 + 0} + 1} = \frac{3}{1 + 1} = \boxed{\frac{3}{2}}$$

Level 3 Radical with Negative Infinity

Evaluate both limits:

$\lim_{x \to +\infty} \frac{\sqrt{4x^2 + x}}{x - 1}$
$\lim_{x \to -\infty} \frac{\sqrt{4x^2 + x}}{x - 1}$

Thought Process

For (a), $x > 0$ so $\sqrt{x^2} = x$. For (b), $x < 0$ so $\sqrt{x^2} = -x$. This sign difference will give different answers!

Show Answer

(a) As $x \to +\infty$: (Here $x > 0$, so $\sqrt{x^2} = x$)

Divide by $x$: $$\frac{\sqrt{4x^2 + x}}{x - 1} = \frac{\sqrt{x^2(4 + \frac{1}{x})}}{x(1 - \frac{1}{x})} = \frac{x\sqrt{4 + \frac{1}{x}}}{x(1 - \frac{1}{x})} = \frac{\sqrt{4 + \frac{1}{x}}}{1 - \frac{1}{x}}$$

As $x \to +\infty$: $$\lim_{x \to +\infty} \frac{\sqrt{4 + \frac{1}{x}}}{1 - \frac{1}{x}} = \frac{\sqrt{4}}{1} = \boxed{2}$$

(b) As $x \to -\infty$: (Here $x < 0$, so $\sqrt{x^2} = -x$)

When we factor $x^2$ out of the radical and take the square root: $$\sqrt{4x^2 + x} = \sqrt{x^2\left(4 + \frac{1}{x}\right)} = \vert x\vert \sqrt{4 + \frac{1}{x}} = -x\sqrt{4 + \frac{1}{x}}$$

(The $-x$ appears because $\vert x\vert = -x$ when $x < 0$!)

So: $$\frac{\sqrt{4x^2 + x}}{x - 1} = \frac{-x\sqrt{4 + \frac{1}{x}}}{x - 1} = \frac{-x\sqrt{4 + \frac{1}{x}}}{x(1 - \frac{1}{x})} = \frac{-\sqrt{4 + \frac{1}{x}}}{1 - \frac{1}{x}}$$

As $x \to -\infty$: $$\lim_{x \to -\infty} \frac{-\sqrt{4 + \frac{1}{x}}}{1 - \frac{1}{x}} = \frac{-\sqrt{4}}{1} = \boxed{-2}$$

Different horizontal asymptotes: $y = 2$ on the right, $y = -2$ on the left.

Level 4 Combined Techniques

Evaluate $\lim_{x \to \infty} \left(\sqrt{x^2 + x} - \sqrt{x^2 - x}\right)$

Thought Process

This is a difference of two radicals, both approaching infinity. Multiply by the conjugate of the entire expression: $\frac{\sqrt{x^2 + x} + \sqrt{x^2 - x}}{\sqrt{x^2 + x} + \sqrt{x^2 - x}}$

Show Answer

Let $A = \sqrt{x^2 + x}$ and $B = \sqrt{x^2 - x}$. We want $\lim_{x \to \infty} (A - B)$.

Multiply by conjugate: $$A - B = \frac{(A - B)(A + B)}{A + B} = \frac{A^2 - B^2}{A + B}$$

Compute numerator: $$A^2 - B^2 = (x^2 + x) - (x^2 - x) = 2x$$

So: $$A - B = \frac{2x}{\sqrt{x^2 + x} + \sqrt{x^2 - x}}$$

For $x > 0$, divide by $x$: $$= \frac{2}{\frac{\sqrt{x^2 + x}}{x} + \frac{\sqrt{x^2 - x}}{x}} = \frac{2}{\sqrt{1 + \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}}$$

As $x \to \infty$: $$\lim_{x \to \infty} \frac{2}{\sqrt{1 + \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}} = \frac{2}{\sqrt{1} + \sqrt{1}} = \frac{2}{2} = \boxed{1}$$

Level 5 Determining Existence of Limits

For each function, determine whether $\lim_{x \to \infty} f(x)$ exists. If it does, find it. If not, explain why.

$f(x) = \sin(x) + \cos(x)$
$f(x) = \frac{\sin(x)}{x}$
$f(x) = e^{-x}\sin(x)$
$f(x) = \sin\left(\frac{1}{x}\right)$

Thought Process

For each:

Does the function oscillate with constant amplitude? → No limit
Does the oscillation decay? → Squeeze theorem may apply
Is there a substitution that helps?

Show Answer

(a) $f(x) = \sin(x) + \cos(x)$

This can be rewritten as $\sqrt{2}\sin\left(x + \frac{\pi}{4}\right)$, which oscillates between $-\sqrt{2}$ and $\sqrt{2}$ forever.

Limit does not exist (oscillation without decay).

(b) $f(x) = \frac{\sin(x)}{x}$

We have $-1 \leq \sin(x) \leq 1$, so: $$-\frac{1}{x} \leq \frac{\sin(x)}{x} \leq \frac{1}{x}$$

As $x \to \infty$, both $-\frac{1}{x}$ and $\frac{1}{x}$ approach $0$.

By the Squeeze Theorem: $\lim_{x \to \infty} \frac{\sin(x)}{x} = \boxed{0}$

(c) $f(x) = e^{-x}\sin(x)$

We have $-1 \leq \sin(x) \leq 1$, so: $$-e^{-x} \leq e^{-x}\sin(x) \leq e^{-x}$$

As $x \to \infty$, $e^{-x} \to 0$ and $-e^{-x} \to 0$.

By the Squeeze Theorem: $\lim_{x \to \infty} e^{-x}\sin(x) = \boxed{0}$

(The decaying exponential "kills" the oscillation.)

(d) $f(x) = \sin\left(\frac{1}{x}\right)$

Let $t = \frac{1}{x}$. As $x \to \infty$, $t \to 0^+$.

$$\lim_{x \to \infty} \sin\left(\frac{1}{x}\right) = \lim_{t \to 0^+} \sin(t) = \sin(0) = \boxed{0}$$

Summary: | Function | Limit Exists? | Value/Reason | |----------|---------------|--------------| | $\sin x + \cos x$ | No | Oscillates with constant amplitude | | $\frac{\sin x}{x}$ | Yes | $0$ (Squeeze) | | $e^{-x}\sin x$ | Yes | $0$ (Squeeze, exponential decay) | | $\sin(1/x)$ | Yes | $0$ (substitution) |

Mastery Checklist

[ ] I can recognize when conjugate multiplication is needed ($\infty - \infty$ with radicals)
[ ] I correctly handle $\sqrt{x^2} = \vert x\vert = -x$ when $x < 0$
[ ] I know that $x \to -\infty$ can give different horizontal asymptotes than $x \to +\infty$
[ ] I can use substitution for limits involving compositions like $\sin(1/x)$
[ ] I can identify when a limit does not exist due to oscillation
[ ] I can apply the Squeeze Theorem for bounded oscillating functions

Mental Model

The Sign Trap Diagram:

When dealing with $\sqrt{x^2}$, picture this:

The number line splits into two regions:

          x < 0                    x > 0
    ←───────────────|───────────────→
          √(x²) = -x      √(x²) = x
              ↑               ↑
         "flip sign"     "keep sign"

The square root always produces a positive result, but $x$ itself is negative on the left side. So $\sqrt{x^2} = \vert x\vert $, which equals $-x$ (a positive number!) when $x$ is negative.

Think of it as: "The square root is always positive, so match its sign to a positive quantity."

Connections

Looking back:

Rational Function Limits handles the basic divide-by-highest-power technique
Indeterminate Forms introduces algebraic tricks like conjugates

Looking ahead:

Curve Sketching combines these limits with derivatives for complete graphs
L'Hôpital's Rule (Calculus II) offers an alternative approach to some of these limits
Improper Integrals require these techniques for evaluating $\int_1^\infty$ types

Real-world connections:

Damped oscillations (like in shock absorbers) involve $e^{-x}\sin(x)$ type behavior
Signal processing uses limits of compositions to analyze frequency response
The $\sqrt{x^2} = \vert x\vert $ issue arises in physics when position/velocity have sign conventions

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Last updated: 2026-01-22