Problems Based on Direct Substitution

In calculus, one of the fundamental concepts is the limit of a function. The limit describes the behavior of a function as its argument approaches a particular point. One of the simplest methods to evaluate limits is direct substitution, which is applicable when the function is continuous at the point of interest.

Understanding Direct Substitution

Direct substitution involves plugging the point of interest directly into the function. If the function is continuous at that point and the substitution yields a finite number, then the limit is equal to that number.

When to Use Direct Substitution

Direct substitution can be used when:

• The function is polynomial.
• The function is rational and the denominator is not zero at the point of interest.
• The function is continuous at the point where the limit is being evaluated.

Table of Differences and Important Points

Aspect	Direct Substitution	Other Methods (e.g., Factoring, Rationalization)
Applicability	Functions that are continuous at the point of interest	Functions that are not continuous or have indeterminate forms at the point of interest
Complexity	Simple and quick	May involve more complex algebraic manipulations
Result	Directly gives the limit if applicable	May require additional steps to resolve indeterminate forms
Example	$\lim_{x \to 2} (3x + 1)$	$\lim_{x \to 0} \frac{\sin(x)}{x}$

Formulas and Direct Substitution

For a continuous function $f(x)$, the limit as $x$ approaches $a$ is given by:

$$ \lim_{x \to a} f(x) = f(a) $$

This is the essence of direct substitution: if you can substitute $a$ into $f(x)$ without any issues, then the limit is simply $f(a)$.

Examples to Explain Important Points

Example 1: Polynomial Function

Evaluate the limit $\lim_{x \to 3} (2x^2 - 5x + 1)$.

Solution:

Since the function is a polynomial, it is continuous everywhere. We can use direct substitution:

$$ \lim_{x \to 3} (2x^2 - 5x + 1) = 2(3)^2 - 5(3) + 1 = 18 - 15 + 1 = 4 $$

Example 2: Rational Function

Evaluate the limit $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.

Solution:

Direct substitution would give us $\frac{0}{0}$, which is an indeterminate form. Therefore, we cannot use direct substitution in this case. Instead, we factor and simplify:

$$ \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{x \to 2} (x + 2) = 4 $$

Example 3: Continuous Function at a Point

Evaluate the limit $\lim_{x \to \pi} \sin(x)$.

Solution:

The sine function is continuous everywhere. We can use direct substitution:

$$ \lim_{x \to \pi} \sin(x) = \sin(\pi) = 0 $$

Example 4: Function with a Hole

Evaluate the limit $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$.

Solution:

Direct substitution would give us $\frac{0}{0}$, which is an indeterminate form. However, we can factor and simplify:

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

In this case, even though the function is not defined at $x = 1$, it is continuous at that point if we consider the hole in the graph. After factoring, we can use direct substitution on the simplified function.

Conclusion

Direct substitution is a powerful tool for evaluating limits when the function is continuous at the point of interest. It is a straightforward method that can save time and effort. However, when faced with indeterminate forms or discontinuities, other methods such as factoring, rationalization, or L'Hôpital's rule may be necessary. Understanding when and how to apply direct substitution is crucial for solving limit problems efficiently.