Determining Limits using Algebraic Manipulation

In calculus, a limit is a fundamental concept that describes the value that a function approaches as the input approaches a certain value. The Limits help us understand the behavior of the functions at points where they might not be explicitly defined. For example, the limit of the function f(x) as x approaches a value a is denoted as:

\lim\limits_{x \to a} f(x)

When evaluating limits, we sometimes encounter indeterminate. The most common indeterminate forms are generated from the limits of the ratio of functions,\frac{f(x)}{g(x)}, when both functions evaluate either to 0 or ∞ generate the limits of the form \frac{0}{0} or\frac{\infty}{\infty}. Other indeterminate forms include 0 x ∞, ∞ - ∞, 0^∞ , etc.

Algebraic techniques help simplify expressions and eliminate indeterminate forms. The main methods include:

1. Limits by Factoring

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\limits_{x \to 1}\frac{x^2 - 1}{x - 1}

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

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

⇒\lim\limits_{x \to 1}\frac{(x - 1)(x + 1)}{x - 1}

⇒\lim\limits_{x \to 1}(x + 1)

⇒\lim\limits_{x \to 1}(1 + 1)=2

2. Limits by Rationalization

This method is used when expressions involve radicals and lead to indeterminate forms such as\infty - \infty\ \text{or}\ \frac{0}{0}.

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

Notice that this generates the indeterminate form of ∞ - ∞.

In such cases, we rationalize the expression.

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

⇒\lim_{x \to \infty} (\sqrt{x^2 + x + 1} - \sqrt{x^2 + 1})\frac{(\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1})}{(\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1})}

⇒\lim_{x \to \infty} \frac{(x^2 + x + 1 - (x^2 + 1))}{(\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1})}

⇒\lim_{x \to \infty} \frac{x}{(\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1})}

⇒\lim_{x \to \infty} \frac{x}{(x\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + \sqrt{1 + \frac{1}{x^2}})}

⇒\lim_{x \to \infty} \frac{1}{(\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + \sqrt{1 + \frac{1}{x^2}})}

⇒\frac{1}{(\sqrt{1 + \frac{1}{\infty} + \frac{1}{\infty^2}} + \sqrt{1 + \frac{1}{\infty^2}})}

⇒\frac{1}{2}

3. Trigonometric limits using Identities

When trigonometric expressions produce indeterminate forms, identities help simplify them.

Pythagorean Identities:

sin²x + cos²x = 1
1 + tan²x = sec²x
1 + cot²x = cosec²x

Double Angle Identities: These identities are useful in transforming expressions

cos2x=2cos²x−1 = 1−2sin²x
sin⁡2x = 2sin⁡xcos⁡x

Sample Problems

Question 1: Find out the following limit. \lim_{x \to 1}\frac{x^2 -3x + 2}{x^2 - 1}

\lim_{x \to 1}\frac{x^2 -3x + 2}{x^2 - 1}
This limit is of the 0/0 form. Factorization can be used here.
\lim_{x \to 1}\frac{x^2 -3x + 2}{x^2 - 1}
⇒\lim_{x \to 1}\frac{x^2 -2x -x + 2}{x^2 - 1^2}
⇒\lim_{x \to 1}\frac{x(x -2) -1(x - 2)}{(x- 1)(x + 1)}
⇒\lim_{x \to 1}\frac{(x - 1)(x -2)}{(x- 1)(x + 1)}
⇒\lim_{x \to 1}\frac{(x -2)}{(x + 1)}
⇒\lim_{x \to 1}\frac{(1 -2)}{(1 + 1)}
⇒\frac{-1}{2}

Question 2: Find out the following limit. \lim_{x \to 2}\frac{x^2 -5x + 6}{x^2 -3x +2}

\lim_{x \to 2}\frac{x^2 -5x + 6}{x^2 -3x +2}
This limit is of the 0/0 form. Factorization can be used here.
\lim_{x \to 2}\frac{x^2 -5x + 6}{x^2 -3x +2}
⇒\lim_{x \to 2}\frac{x^2 -3x -2x + 6}{x^2 -x -2x +2}
⇒\lim_{x \to 2}\frac{(x -3)(x -2)}{(x -1)(x-2)}
⇒\lim_{x \to 2}\frac{(x -3)}{(x -1)}
⇒\lim_{x \to 2}\frac{(2 -3)}{(2 -1)}
⇒ -1

Question 3: Find out the following limit. \lim_{x \to 9}\frac{\sqrt{x} - 3}{x -9}

Substitute x = 9:
\frac{\sqrt{9} - 3}{9 - 9} = \frac{3 - 3}{0} = \frac{0}{0}
This is an indeterminate form, so we need to simplify.
Simplify using conjugate:
Multiply numerator and denominator by the conjugate of the numerator, √x + 3:
\dfrac{\sqrt{x} - 3}{x - 9} \times \dfrac{\sqrt{x} + 3}{\sqrt{x} + 3} \\= \dfrac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)} \\= \dfrac{x - 9}{(x - 9)(\sqrt{x} + 3)} \\ = \dfrac{1}{\sqrt{x} + 3}
Evaluate the simplified limit:
\lim_{x \to 9} \frac{1}{\sqrt{x} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}

Question 4: Find out the limit of, lim_{x\to2}\frac{x-2}{(\sqrt{x+2}-2)}

Putting the limit x tends to 2 to see the value obtained,
lim_{x\to2}\frac{x-2}{(\sqrt{x+2}-2)}\\=\frac{0}{0}\\
= Undefined.
As, it is clear that the answer is undefined, rationalize the denominator,
lim_{x\to2}\frac{x-2}{(\sqrt{x+2}-2)}\\=lim_{x\to2}\frac{(x-2)(\sqrt{x+2}+2)}{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}\\=lim_{x\to2}\frac{(x-2)(\sqrt{x+2}+2)}{(x-2)}\\=lim_{x\to2}\sqrt{x+2}+2\\=2+2=4

Practice Problems

Problem 1: Find the limit: \lim_{x \to 3} \frac{x^2 - 9}{x - 3}

Problem 2: Find the limit: \lim_{x \to 2} \frac{x^2 - 4}{x - 2}

Problem 3: Find the limit: \lim_{x \to 0} \frac{x^2 + 2x}{x}

Problem 4: Find the limit: \lim_{x \to 1} \frac{x^3 - 1}{x - 1}

Problem 5: Find the limit: \lim_{x \to -1} \frac{x^2 - 1}{x + 1}

Limits in Calculus
Strategy in Finding Limits
Limits by Direct Substitution
Estimating Limits from Graphs