Solving Infinite Limits

Sahabat Smaridasa - Mathematical Concepts

In this article, we will discuss Infinite Limits. "Infinite limits" can mean either the limit result is infinity (\(\infty\)) or the variable approaches infinity (\(x \to \infty\)). Please also refer to the materials on "Definition of Function Limits" and "Solving Algebraic Function Limits". Here, we will focus on algebraic functions. Infinite limits for trigonometric functions will be covered separately.

Limits Whose Result is Infinity

A limit results in infinity (\(\infty\)) if the output grows without bound. This often occurs when the denominator approaches zero ( \(\frac{1}{0} = \infty\) ).

Basic Theory:

\[ \lim_{x \to 0^+} \frac{1}{x^n} = +\infty, \quad \lim_{x \to 0^-} \frac{1}{x^n} = \begin{cases} +\infty, & n \text{ even} \\ -\infty, & n \text{ odd} \end{cases} \]

Note: If the exponent \(n\) is even, the result is always positive.

📘 Example 1 :

Find \(\displaystyle \lim_{x \to 2} \frac{1}{(x-2)^2}\).

Solution:

As \(x \to 2\), \((x-2)^2 \to 0^+\). Therefore, \(\frac{1}{(x-2)^2} \to +\infty\).

Thus, \(\displaystyle \lim_{x \to 2} \frac{1}{(x-2)^2} = \infty\).

Grafik \( f(x) = \frac{1}{(x-2)^2} \) menunjukkan asimtot vertikal di \( x = 2 \). Nilai fungsi membesar tak hingga ketika \( x \) mendekati 2.

📘 Example 2 :

Find the following limits:

a) \(\displaystyle \lim_{x \to 5^+} \frac{x+2}{(x-5)^5}\)    b) \(\displaystyle \lim_{x \to 3^-} \frac{2}{(x-3)^8}\)    c) \(\displaystyle \lim_{x \to 3^-} \frac{2}{(x-3)^7}\)

Solution:

a) As \(x \to 5^+\), \(x-5 \to 0^+\), so \((x-5)^5 \to 0^+\). The numerator \(x+2 \to 7\). Hence, \(\frac{7}{0^+} = +\infty\).

b) As \(x \to 3^-\), \(x-3 \to 0^-\). Since the exponent 8 is even, \((x-3)^8 \to 0^+\). Then \(\frac{2}{0^+} = +\infty\).

c) As \(x \to 3^-\), \(x-3 \to 0^-\). Since the exponent 7 is odd, \((x-3)^7 \to 0^-\). Then \(\frac{2}{0^-} = -\infty\).

Solving Limits at Infinity (\(x \to \infty\))

To evaluate limits as \(x \to \infty\), we use the basic limit: \(\displaystyle \lim_{x \to \infty} \frac{a}{x^n} = 0\) for real \(a\) and natural \(n\). We transform the function into a fraction, divide by the highest power of \(x\), and apply the basic limit.

📘 Example 3 : Limits at Infinity (General Method)

Evaluate the following limits:

a) \(\displaystyle \lim_{x \to \infty} \frac{2x^3 + 3x^2 + 5}{5x^3 - 4x + 1}\)    b) \(\displaystyle \lim_{x \to \infty} \frac{-2x^2 - 5}{5x^8 - 4x + 3}\)    c) \(\displaystyle \lim_{x \to \infty} \frac{x^5 - 2x^3 + 5x - 1}{3x^2 - 4x + 1}\)

d) \(\displaystyle \lim_{x \to \infty} \frac{\sqrt{4x^2 + 2x - 3} - \sqrt{4x^2 - x + 3}}{\sqrt{9x^2 + 2x - 7}}\)    e) \(\displaystyle \lim_{x \to \infty} \left( \sqrt{4x^2 + 2x - 3} - \sqrt{4x^2 - x + 3} \right)\)

Solution (using highest power division):

a) Divide numerator and denominator by \(x^3\): \[ \lim_{x \to \infty} \frac{2 + \frac{3}{x} + \frac{5}{x^3}}{5 - \frac{4}{x^2} + \frac{1}{x^3}} = \frac{2}{5}. \]

b) Highest power in denominator is \(x^8\). Divide: \[ \lim_{x \to \infty} \frac{-\frac{2}{x^6} - \frac{5}{x^8}}{5 - \frac{4}{x^7} + \frac{3}{x^8}} = \frac{0}{5} = 0. \]

c) Highest power in numerator is \(x^5\), denominator \(x^2\). As \(x \to \infty\), numerator dominates → \(\infty\). So limit = \(\infty\).

d) For the numerator, factor \(x^2\) inside the square roots: \(\sqrt{4x^2(1 + \frac{2}{4x} - \frac{3}{4x^2})} = x\sqrt{4 + \frac{2}{x} - \frac{3}{x^2}}\), similarly \(\sqrt{4x^2 - x + 3} = x\sqrt{4 - \frac{1}{x} + \frac{3}{x^2}}\). Denominator: \(\sqrt{9x^2 + 2x - 7} = x\sqrt{9 + \frac{2}{x} - \frac{7}{x^2}}\). Then: \[ \lim_{x \to \infty} \frac{x\left(\sqrt{4 + \frac{2}{x} - \frac{3}{x^2}} - \sqrt{4 - \frac{1}{x} + \frac{3}{x^2}}\right)}{x\sqrt{9 + \frac{2}{x} - \frac{7}{x^2}}} = \frac{\sqrt{4} - \sqrt{4}}{\sqrt{9}} = \frac{2-2}{3} = 0. \]

e) Multiply by conjugate: \[ \lim_{x \to \infty} \frac{(4x^2+2x-3) - (4x^2 - x + 3)}{\sqrt{4x^2+2x-3} + \sqrt{4x^2 - x + 3}} = \lim_{x \to \infty} \frac{3x - 6}{2x\sqrt{1 + \frac{2}{4x} - \frac{3}{4x^2}} + 2x\sqrt{1 - \frac{1}{4x} + \frac{3}{4x^2}}} = \frac{3}{2+2} = \frac{3}{4}. \]

Practical Rules for Limits at Infinity

For rational functions: \(\displaystyle \lim_{x \to \infty} \frac{a_n x^n + \dots}{b_m x^m + \dots} = \begin{cases} 0, & n < m \\ \frac{a_n}{b_m}, & n = m \\ \infty \text{ or } -\infty, & n > m \end{cases}\)

For the form \(\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q}\): \[ \lim_{x \to \infty} \left( \sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q} \right) = \frac{b-p}{2\sqrt{a}} \]

📘 Example 4 : Using the Shortcut (Same \(a\))

Use the shortcut to solve:

a) \(\displaystyle \lim_{x \to \infty} \frac{2x^3 + 3x^2 + 5}{5x^3 - 4x + 1}\)    b) \(\displaystyle \lim_{x \to \infty} \frac{-2x^2 - 5}{5x^8 - 4x + 3}\)    c) \(\displaystyle \lim_{x \to \infty} \frac{x^5 - 2x^3 + 5x - 1}{3x^2 - 4x + 1}\)

d) \(\displaystyle \lim_{x \to \infty} \frac{\sqrt{4x^2 + 2x - 3} - \sqrt{4x^2 - x + 3}}{\sqrt{9x^2 + 2x - 7}}\)    e) \(\displaystyle \lim_{x \to \infty} \left( \sqrt{4x^2 + 2x - 3} - \sqrt{4x^2 - x + 3} \right)\)

Solution using practical rules:

a) Degree numerator = degree denominator = 3 → limit = coefficient ratio = \(\frac{2}{5}\).

b) Degree numerator (2) < degree denominator (8) → limit = 0.

c) Degree numerator (5) > degree denominator (2) → limit = \(\infty\).

d) For large \(x\), numerator \(\approx 2x - 2x = 0\), denominator \(\approx 3x\). More precisely, the limit is 0.

e) Using the formula \(\frac{b-p}{2\sqrt{a}}\): here \(a=4\), \(b=2\), \(p=-1\) → \(\frac{2 - (-1)}{2\sqrt{4}} = \frac{3}{4}\).

📘 Example 5 : More Complex Forms at Infinity

Find the following limits:

a) \(\displaystyle \lim_{x \to \infty} \left( \sqrt{x^2 - 5x} - (x + 2) \right)\)

b) \(\displaystyle \lim_{x \to \infty} \left( 2x - 3 - \sqrt{4x^2 + x - 7} \right)\)

c) \(\displaystyle \lim_{x \to \infty} \left( \frac{5x + 3}{5x^2 + 2} - 7 \right)\)

Solution:

a) Rewrite \(x+2 = \sqrt{(x+2)^2}\). Then: \[ \lim_{x \to \infty} \left( \sqrt{x^2 - 5x} - \sqrt{x^2 + 4x + 4} \right) = \frac{b-p}{2\sqrt{a}} = \frac{-5 - 4}{2\sqrt{1}} = \frac{-9}{2}. \]

b) Write \(2x-3 = \sqrt{(2x-3)^2} = \sqrt{4x^2 -12x + 9}\). Then: \[ \lim_{x \to \infty} \left( \sqrt{4x^2 -12x + 9} - \sqrt{4x^2 + x - 7} \right) = \frac{-12 - 1}{2\sqrt{4}} = \frac{-13}{4}. \]

c) \(\displaystyle \lim_{x \to \infty} \left( \frac{5x+3}{5x^2+2} - 7 \right) = 0 - 7 = -7\). Because \(\frac{5x+3}{5x^2+2} \to 0\).