How to Determine Limits Using Algebraic Manipulation?

Some limits may not be defined because they are divisible by \(0\). We can usually define such a limit by algebraic manipulations. We will learn this in this article.

Related Topics

• How to Define Limits Analytically Using Correct Notation
• How to Estimate Limit Values from the Graph
• Properties of Limits

A step-by-step guide to determining limits using algebraic manipulation

In the limits, these algebraic approaches can assist in avoiding indeterminate forms. Limits can be evaluated in a variety of ways, including factoring and reasoning.

Other approaches, such as the Pythagorean identity or the trigonometric limit utilizing the double angle identity, can assist us in solving these limits in the case of trigonometric functions.

Factoring-In Limits

• Usually, in the ratio functions consisting of polynomials, the indeterminate form stems from one of the factors occurring in the expression. For example, in the function \(f(x)\) given below, the indeterminate form is due to the factor \((x – 1)\).

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

In such cases, we factorize both the polynomials such that the common factor cancels out.

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

\(= lim_{x\to 2}\frac{(x-1)(x+1)}{(x-1)}\)

\(=lim_{x\to 2}(x+1)\)

\(= lim_{x\to 2}(2+1)\)

\(=3\)

Rationalization-Induced Limits

• The function is rationalized in this approach to cope with indeterminate forms. For example, \(lim_{x\to ∞}\sqrt{x^2+x+2}-\sqrt{x^2+2}\)

In such cases, we rationalize the expression.

\(lim_{x\to ∞}\sqrt{x^2+x+2}-\sqrt{x^2+2}\)

\(=lim _{x\to \infty }\left(\sqrt{x^2+x+2}-\sqrt{x^2+2}\right)\times\frac{\left(\sqrt{x^2+x+2}\:+\:\sqrt{x^2+2}\right)}{\left(\sqrt{x^2+x+2}\:+\:\sqrt{x^2+2}\right)}\)

\(=\lim _{x\to \infty }\left(\frac{x^2+x+2-\left(x^2+2\right)}{\sqrt{x^2+x+2}\:+\:\sqrt{x^2+2}}\right)\)

\(=lim _{x\to \infty }\left(\frac{x}{\sqrt{x^2+x+2}\:+\:\sqrt{x^2+2}}\right)\)

\(=lim _{x\to \infty }\left(\frac{x}{x\sqrt{1+\frac{1}{x}+\frac{2}{x^2}}\:+\:\sqrt{1+\frac{2}{x^2}}}\right)\)

\(=lim _{x\to \infty }\left(\frac{1}{\sqrt{1+\frac{1}{x}+\frac{2}{x^2}}\:+\:\sqrt{1+\frac{2}{x^2}}}\right)\)

\(=lim _{x\to \infty }\left(\frac{1}{\sqrt{1+\frac{1}{ \infty }+\frac{2}{ \infty ^2}}\:+\:\sqrt{1+\frac{2}{ \infty ^2}}}\right)\)

\(=\frac{1}{2}\)

Determining Limits Using Algebraic Manipulation – Example 1:

calculate the following limit. \(lim _{x\to 4}\left(\frac{x^2-5x+6}{x^2-3x+2}\right)\)

\(lim _{x\to 4}\left(\frac{x^2-5x+6}{x^2-3x+2}\right)\) \(=lim_{x\to 4}\frac{(x-2)(x-3)}{(x-2)(x-1)}\)

\(=lim_{x\to 4}\frac{(x-3)}{(x-1)}\)

\(=lim_{x\to 4}\frac{(4-3)}{(4-1)}\)

\(=\frac{1}{3}\)

Determining Limits Using Algebraic Manipulation – Example 2:

calculate the following limit. \(lim _{x\to 4}\left(\frac{\sqrt{x}-3}{x-9}\right)\)

\(lim _{x\to 4}\left(\frac{\sqrt{x}-3}{x-9}\right)\)

\(=lim _{x\to 4}\left(\frac{\sqrt{x}-3}{x-9}\right) \times \frac{\sqrt{x}+3}{\sqrt{x}+3}\)

\(=lim _{x\to 4}\left(\frac{\left(x-9\right)}{\left(x-9\right)\left(\sqrt{x}+3\right)}\right)\)

\(=lim _{x\to 4}\left(\frac{1}{\left(\sqrt{x}+3\right)}\right)\)

\(=lim _{x\to 4}\left(\frac{1}{\left(\sqrt{4}+3\right)}\right)=lim _{x\to 4}\left(\frac{1}{\left(2+3\right)}\right)\)

\(=\frac{1}{5}\)

Exercises for Determining Limits Using Algebraic Manipulation

Calculate the value of the following limit.

1. \(\color{blue}{lim _{x\to 4}\left(\frac{\sqrt{3x+4}-5}{x-5}\right)}\)
2. \(\color{blue}{lim _{x\to \infty }\left(\frac{x^2-3x+3}{x^3+4x+2}\right)}\)
3. \(\color{blue}{lim _{x\to 2}\left(\frac{x^2-3x-28}{x^2-6x-7}\right)}\)
4. \(\color{blue}{lim _{x\to 5}\left(\frac{3x^2+12x}{x+4}\right)}\)
5. \(\color{blue}{lim _{x\to 0}\left(\frac{\sqrt[4]{x^4+8}-2}{x^4}\right)}\)

1. \(\color{blue}{1}\)
2. \(\color{blue}{0}\)
3. \(\color{blue}{2}\)
4. \(\color{blue}{15}\)
5. \(\color{blue}{-\infty}\)