#Substituting to Find Limits

#Table of Contents

1. Introduction to Substitution
2. Substitution for Continuous Functions
3. Substitution for Piecewise Functions
4. Simplifying to Find Limits
5. Multiplying by Conjugates to Find Limits
6. Multiplying by Reciprocals to Find Limits at Infinity
7. Practice Questions
8. Glossary
9. Summary and Key Takeaways

---

#Introduction to Substitution

Substitution is a straightforward method to evaluate limits by directly replacing the variable with the value it approaches. However, this method is often useful only for continuous functions. Below, we explore various techniques to find limits using substitution and other methods when substitution alone does not suffice.

---

#Substitution for Continuous Functions

#How can I use substitution to find limits?

If a function $f$ is continuous on an open interval, and if $a$ is a number contained in that interval, then:

$$\underset{x\to a}{\mathrm{lim}}f(x)=f(a)$$

This means for a continuous function, the limit at a point is equal to the value of the function at that point.

This idea can be extended to a point that is **on or 'just outside'** the border of an interval where a function (or part of a function) is defined. This is especially useful for **one-sided limits** at the 'joins' of piecewise functions.

#Example: Piecewise Function

Consider the function $f$ defined by:

$$f(x) = \begin{cases} x-7, & x \leq 3 \\ x^2, & x > 3 \end{cases}$$

• Both $x-7$ and $x^2$ are continuous functions on all real numbers.
• To find the one-sided limits:
  • $$\underset{x\to 3^{-}}{\mathrm{lim}}f(x) = 3-7 = -4$$
  • $$\underset{x\to 3^{+}}{\mathrm{lim}}f(x) = 3^2 = 9$$

#Worked Example

Let $f$ be the function defined by:

$$f(x) = \begin{cases} x^3 - 3x + 4, & x < -2 \\ x^2 - 2, & x \geq -2 \end{cases}$$

Find:

(a) $\underset{x\to 4}{\mathrm{lim}}f(x)$

Answer:

Since $x=4$ belongs to the $x^2 - 2$ piece of the function, and $x^2 - 2$ is continuous on the open interval $(-2, \infty)$:

$$\underset{x\to 4}{\mathrm{lim}}f(x) = f(4) = 4^2 - 2 = 14$$

(b) $\underset{x\to -2}{\mathrm{lim}}f(x)$

For this limit to exist, the limits from the left and right must both exist and be equal.

• $$\underset{x\to -2^{-}}{\mathrm{lim}}f(x) = (-2)^3 - 3(-2) + 4 = -8 + 6 + 4 = 2$$
• $$\underset{x\to -2^{+}}{\mathrm{lim}}f(x) = (-2)^2 - 2 = 4 - 2 = 2$$

Since both one-sided limits are equal:

$$\underset{x\to -2}{\mathrm{lim}}f(x) = 2$$

---

#Simplifying to Find Limits

#How can I simplify functions to find limits?

Sometimes, evaluating a limit by substitution gives an answer that is not defined (e.g., returning a value like $\frac{0}{0}$). In these cases, the equation of the function can often be simplified by factorizing and canceling common factors. After simplification, substitution may provide a well-defined value for the limit.

#Example: Simplification

Let $f$ be the function defined by:

$$f(x) = \frac{(x-1)(x-3)}{x(x-1)}$$

• The function is undefined at $x=1$ because $f(1) = \frac{0}{0}$.
• To find the limit at $x=1$, we can cancel the common factor $(x-1)$:

$$\underset{x\to 1}{\mathrm{lim}}f(x) = \underset{x\to 1}{\mathrm{lim}} \frac{(x-3)}{x} = \frac{1-3}{1} = -2$$

#Worked Example

Let $f$ be the function defined by:

$$f(x) = \frac{x^2 + 2x}{x^2 - x - 6}$$

Find $\underset{x\to -2}{\mathrm{lim}}f(x)$.

Answer:

Attempting to evaluate this limit by substitution doesn't work:

$$\frac{(-2)^2 + 2(-2)}{(-2)^2 - (-2) - 6} = \frac{4 - 4}{4 + 2 - 6} = \frac{0}{0}$$

Instead, factorize the numerator and denominator, simplify the fraction, and use substitution to evaluate the limit in the simplified form:

$$\frac{x^2 + 2x}{x^2 - x - 6} = \frac{x(x+2)}{(x+2)(x-3)} = \frac{x}{x-3}$$

Now use substitution:

$$\frac{-2}{-2-3} = \frac{-2}{-5} = \frac{2}{5}$$

$$\underset{x\to -2}{\mathrm{lim}}f(x) = \frac{2}{5}$$

---

#Multiplying by Conjugates to Find Limits

#How can I use multiplying by conjugates to help find limits?

A quotient involving surds (i.e., square roots) can sometimes be rewritten by multiplying it top and bottom by a conjugate of either the numerator or denominator. The conjugate of $\sqrt{a}-\sqrt{b}$ is $\sqrt{a}+\sqrt{b}$.

#Example: Conjugates

Consider the function:

$$\frac{\sqrt{5}-\sqrt{x+5}}{x}$$

By multiplying the top and bottom by the conjugate of the numerator:

$$\frac{\sqrt{5}-\sqrt{x+5}}{x} \cdot \frac{\sqrt{5}+\sqrt{x+5}}{\sqrt{5}+\sqrt{x+5}} = \frac{5 - (x+5)}{x(\sqrt{5}+\sqrt{x+5})} = \frac{-x}{x(\sqrt{5}+\sqrt{x+5})} = \frac{-1}{\sqrt{5}+\sqrt{x+5}}$$

#Worked Example

Let $f$ be the function defined by:

$$f(x) = \frac{x}{\sqrt{3}-\sqrt{x+3}}, \quad x \geq -3$$

Find $\underset{x\to 0}{\mathrm{lim}}f(x)$.

Answer:

Attempting to evaluate this limit by substitution doesn't work:

$$\frac{0}{\sqrt{3}-\sqrt{3}} = \frac{0}{0}$$

Instead, multiply the top and bottom by the conjugate of the denominator:

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

Now use substitution:

$$-\sqrt{3} - \sqrt{0+3} = -\sqrt{3} - \sqrt{3} = -2\sqrt{3}$$

$$\underset{x\to 0}{\mathrm{lim}}f(x) = -2\sqrt{3}$$

---

#Multiplying by Reciprocals to Find Limits at Infinity

#How can I use multiplying by reciprocals to find a limit at infinity?

A quotient involving powers of $x$ can be rewritten by multiplying its top and bottom by the reciprocal of the highest power of $x$ in the numerator or denominator. This can be used to find a limit at infinity because if $k$ is any constant and $n$ is a positive integer, then:

$$\underset{x\to \pm \infty }{\mathrm{lim}}\frac{k}{x^n}=0$$

#Example: Reciprocals

Consider the function:

$$\frac{7x^2-4x+13}{2x^2+3x-5}$$

Multiply the top and bottom by the reciprocal of the highest power of $x$ in the denominator:

$$\frac{7x^2-4x+13}{2x^2+3x-5} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} = \frac{7 - \frac{4}{x} + \frac{13}{x^2}}{2 + \frac{3}{x} - \frac{5}{x^2}}$$

#Worked Example

Let $f$ be the function defined by:

$$f(x) = \frac{4x^3 - 3x^2 + 1}{5x^3 + 2x + 17}$$

Find $\underset{x\to \infty}{\mathrm{lim}}f(x)$.

Answer:

The numerator and denominator both tend to infinity as $x$ tends to infinity, but $\frac{\infty}{\infty}$ is not defined. Instead, multiply the top and bottom by the reciprocal of the highest power, $x^3$:

$$\frac{4x^3 - 3x^2 + 1}{5x^3 + 2x + 17} \cdot \frac{\frac{1}{x^3}}{\frac{1}{x^3}} = \frac{4 - \frac{3}{x} + \frac{1}{x^3}}{5 + \frac{2}{x^2} + \frac{17}{x^3}}$$

Now take the limit as $x$ tends to infinity:

$$\underset{x\to \infty}{\mathrm{lim}} \frac{4 - 0 + 0}{5 + 0 + 0} = \frac{4}{5}$$

$$\underset{x\to \infty}{\mathrm{lim}}f(x) = \frac{4}{5}$$

---

#Practice Questions

#Substitution to Find Limits

Practice Question

Find $\underset{x\to 2}{\mathrm{lim}}f(x)$ for $f(x) = 3x + 1$.

Practice Question

Find $\underset{x\to -1}{\mathrm{lim}}f(x)$ for $f(x) = \frac{x^2 - 1}{x + 1}$.

#Simplifying to Find Limits

Practice Question

Find $\underset{x\to 1}{\mathrm{lim}}f(x)$ for $f(x) = \frac{x^2 - 1}{x - 1}$.

Practice Question

Find $\underset{x\to 3}{\mathrm{lim}}f(x)$ for $f(x) = \frac{x^2 - 9}{x - 3}$.

#Multiplying by Conjugates to Find Limits

Practice Question

Find $\underset{x\to 4}{\mathrm{lim}}f(x)$ for $f(x) = \frac{\sqrt{x} - 2}{x - 4}$.

Practice Question

Find $\underset{x\to 1}{\mathrm{lim}}f(x)$ for $f(x) = \frac{\sqrt{x+3} - 2}{x - 1}$.

#Multiplying by Reciprocals to Find Limits at Infinity

Practice Question

Find $\underset{x\to \infty}{\mathrm{lim}}f(x)$ for $f(x) = \frac{2x^2 - 3x + 1}{x^2 + x - 4}$.

Practice Question

Find $\underset{x\to -\infty}{\mathrm{lim}}f(x)$ for $f(x) = \frac{5x^3 - x + 7}{3x^3 + 2x^2 - x}$.

---

#Glossary

• Continuous Function: A function without breaks, jumps, or holes in its graph.
• One-Sided Limit: The value a function approaches as the input approaches from one side (left or right) only.
• Piecewise Function: A function defined by different expressions for different intervals of the domain.
• Factorizing: Breaking down a complex expression into simpler components that, when multiplied together, give the original expression.
• Surds: Expressions containing square roots.
• Conjugate: An expression formed by changing the sign between two terms in a binomial (e.g., the conjugate of $\sqrt{a}-\sqrt{b}$ is $\sqrt{a}+\sqrt{b}$).
• Reciprocal: The multiplicative inverse of a number or expression (e.g., the reciprocal of $x$ is $\frac{1}{x}$).

---

#Summary and Key Takeaways

• Substitution is useful for finding limits of continuous functions.
• For piecewise functions, ensure the left-hand and right-hand limits are equal for the limit to exist.
• Simplification by factorizing and canceling common factors can resolve limits that initially return indeterminate forms like $\frac{0}{0}$.
• Multiplying by conjugates is effective for handling limits involving surds.
• Multiplying by reciprocals helps find limits at infinity by simplifying expressions involving powers of $x$.
• Practice these techniques to handle various limit problems effectively.

---

For more complex limits, consider studying L'Hospital's Rule, which provides another method to evaluate limits that result in indeterminate forms.