Limits Made Easy: How to Evaluate Common Limits (with Worked Solutions)

Limits are the cornerstone of calculus. They describe the behavior of a function as its input approaches a certain value. Understanding limits is crucial because they form the basis for derivatives, integrals, and continuity.

What is a Limit? (Intuitive Explanation)

Imagine you're walking towards a wall. As you get closer and closer, your distance to the wall approaches zero. You might never actually reach the wall, but you can get arbitrarily close.

In mathematics, a limit describes the value that a function "approaches" as the input (say, $x$) gets closer and closer to some number. The function doesn't necessarily have to be defined at that specific number for the limit to exist.

Notation of Limits

The notation for a limit is:

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

This is read as "the limit of $f(x)$ as $x$ approaches $a$ is $L$." It means that as $x$ gets very, very close to $a$ (from both sides, without necessarily being equal to $a$), the value of $f(x)$ gets very, very close to $L$.

How to Evaluate Limits

Here are common techniques for evaluating limits, along with worked solutions.

1. Direct Substitution

This is the easiest method. If $f(x)$ is a "well-behaved" function (like polynomials, rational functions where the denominator is not zero, trigonometric functions, etc.) and $f(a)$ is defined, you can just substitute $a$ into the function.

Example 1: Evaluate $\lim_{x \to 2} (x^2 + 3x - 5)$

Solution: Since $x^2 + 3x - 5$ is a polynomial, it's continuous everywhere. We can use direct substitution.

$$\lim_{x \to 2} (x^2 + 3x - 5) = (2)^2 + 3(2) - 5 = 4 + 6 - 5 = 5$$

Answer: $5$

2. Factoring and Canceling

This method is used when direct substitution results in an indeterminate form like $\frac{0}{0}$. Often, this indicates that there's a common factor in the numerator and denominator that can be canceled out.

Example 2: Evaluate $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$

Solution: If we try direct substitution, we get $\frac{3^2 - 9}{3 - 3} = \frac{0}{0}$, which is an indeterminate form. Factor the numerator ($x^2 - 9 = (x - 3)(x + 3)$):

$$\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}$$

Since $x \to 3$, $x \neq 3$, so we can cancel out the $(x - 3)$ terms:

$$\lim_{x \to 3} (x + 3)$$

Now, use direct substitution:

$$\lim_{x \to 3} (x + 3) = 3 + 3 = 6$$

Answer: $6$

3. Rationalizing

This technique is particularly useful when dealing with limits involving square roots that result in the indeterminate form $\frac{0}{0}$. We multiply the numerator and denominator by the conjugate of the expression containing the square root.

Example 3: Evaluate $\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}$

Solution: Direct substitution gives $\frac{\sqrt{0 + 1} - 1}{0} = \frac{0}{0}$. Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x + 1} + 1$:

$$\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x} \cdot \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} + 1}$$ $$= \lim_{x \to 0} \frac{(x + 1) - 1}{x(\sqrt{x + 1} + 1)}$$ $$= \lim_{x \to 0} \frac{x}{x(\sqrt{x + 1} + 1)}$$

Cancel out $x$ (since $x \to 0$, $x \neq 0$):

$$= \lim_{x \to 0} \frac{1}{\sqrt{x + 1} + 1}$$

Now, use direct substitution:

$$= \frac{1}{\sqrt{0 + 1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$$

Answer: $\frac{1}{2}$

4. Limits at Infinity

These limits describe the behavior of a function as $x$ becomes very large (positive or negative). For rational functions, we often look at the highest power of $x$ in the numerator and denominator.

Example 4: Evaluate $\lim_{x \to \infty} \frac{3x^2 + 2x - 1}{x^2 - 4x + 7}$

Solution: Divide every term in the numerator and denominator by the highest power of $x$ in the denominator, which is $x^2$:

$$\lim_{x \to \infty} \frac{\frac{3x^2}{x^2} + \frac{2x}{x^2} - \frac{1}{x^2}}{\frac{x^2}{x^2} - \frac{4x}{x^2} + \frac{7}{x^2}}$$ $$= \lim_{x \to \infty} \frac{3 + \frac{2}{x} - \frac{1}{x^2}}{1 - \frac{4}{x} + \frac{7}{x^2}}$$

As $x \to \infty$, terms like $\frac{2}{x}$, $\frac{1}{x^2}$, $\frac{4}{x}$, and $\frac{7}{x^2}$ all approach $0$.

$$= \frac{3 + 0 - 0}{1 - 0 + 0} = \frac{3}{1} = 3$$

Answer: $3$

5. Special Trigonometric Limits

There are two very important trigonometric limits that often appear:

1. $\lim_{x \to 0} \frac{\sin x}{x} = 1$
2. $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$

Example 5: Evaluate $\lim_{x \to 0} \frac{\sin(4x)}{x}$

Solution: This limit resembles the special limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$. To make the argument of sine match the denominator, we multiply and divide by $4$:

$$\lim_{x \to 0} \frac{\sin(4x)}{x} = \lim_{x \to 0} \frac{\sin(4x)}{x} \cdot \frac{4}{4}$$ $$= \lim_{x \to 0} 4 \cdot \frac{\sin(4x)}{4x}$$

Let $u = 4x$. As $x \to 0$, $u \to 0$. So, the limit becomes:

$$= 4 \lim_{u \to 0} \frac{\sin u}{u} = 4 \cdot 1 = 4$$

Answer: $4$

Common Pitfalls

• Indeterminate Forms: Always check for $\frac{0}{0}$ or $\frac{\infty}{\infty}$ after direct substitution. These require further algebraic manipulation.
• One-Sided Limits: Be aware that limits can approach from the left ($x \to a^-$) or the right ($x \to a^+$). For a general limit to exist, the left-hand and right-hand limits must be equal.
• Limits of Piecewise Functions: Pay special attention to the point where the function definition changes.

Conclusion

Limits are a fundamental concept in calculus, providing the groundwork for understanding rates of change and accumulation. By mastering direct substitution, factoring, rationalizing, limits at infinity, and special trigonometric limits, you'll be well-equipped to tackle more advanced calculus topics. Practice is key!