Finding Limits Algebraically: Videos & Practice Problems

To find limits of functions, especially rational ones, direct substitution is often effective unless the denominator equals zero. In such cases, factor the numerator and denominator, cancel common factors, and evaluate. For functions with radicals, multiply by the conjugate to simplify. Understanding these methods is crucial for determining limits accurately, ensuring a solid grasp of calculus concepts and their applications in various mathematical contexts.

Finding Limits by Direct Substitution

To find the limit of a function as \( x \) approaches a specific value, one effective method is direct substitution. This approach is particularly useful for many functions, including polynomials and basic root functions, where the limit at a point is equal to the function's value at that point. For instance, if you want to find the limit of a polynomial function like \( 6x^3 + 3x^2 - x + 5 \) as \( x \) approaches 2, you can simply substitute 2 into the function. This results in:

\[6(2^3) + 3(2^2) - 2 + 5 = 6(8) + 3(4) - 2 + 5 = 48 + 12 - 2 + 5 = 63\]

Thus, the limit as \( x \) approaches 2 is 63.

Similarly, for a root function such as \( \sqrt{7x^2 + 4x + 16} \) as \( x \) approaches 0, direct substitution yields:

\[\sqrt{7(0^2) + 4(0) + 16} = \sqrt{16} = 4\]

Therefore, the limit is 4.

When dealing with rational functions, you can also use direct substitution, provided that the denominator does not equal zero at the point of interest. For example, to find the limit of \( \frac{x^2 + 3x + 2}{x + 1} \) as \( x \) approaches 0, first check the denominator:

\[0 + 1 = 1 \quad (\text{not } 0)\]

Since the denominator is not zero, you can substitute directly:

\[\frac{0^2 + 3(0) + 2}{0 + 1} = \frac{2}{1} = 2\]

Thus, the limit as \( x \) approaches 0 is 2.

In summary, for many functions, especially polynomials and basic roots, the limit can be found by simply substituting the value of \( x \) into the function. However, caution is needed with rational functions to ensure the denominator does not equal zero at the point of substitution. This method of direct substitution simplifies the process of finding limits significantly.

Finding Limits by Direct Substitution Example 1

In calculus, understanding limits is crucial, especially when dealing with different types of functions. A key concept to remember is that for constant functions, the limit as x approaches any value is simply the constant itself. For example, when finding the limit of the constant function 7 as x approaches 2, the limit is straightforwardly 7, since the function does not change regardless of the value of x.

When working with polynomial functions, such as \( f(x) = 2x^2 + 3x \), the limit can be found by directly substituting the value of x into the function. For instance, to find the limit as x approaches -1, we substitute -1 into the polynomial:

\[ f(-1) = 2(-1)^2 + 3(-1) = 2(1) - 3 = 2 - 3 = -1 \]

Thus, the limit of this function as x approaches -1 is -1.

Another example involves functions with square roots, such as \( f(x) = \sqrt{3x^2 - 2} \). To find the limit as x approaches 3, we substitute 3 into the function:

\[ f(3) = \sqrt{3(3)^2 - 2} = \sqrt{3 \cdot 9 - 2} = \sqrt{27 - 2} = \sqrt{25} = 5 \]

Therefore, the limit of this function as x approaches 3 is 5.

In summary, when evaluating limits, direct substitution is often the most effective method, especially for polynomial and continuous functions. Understanding these principles will aid in solving more complex limit problems in calculus.

Find the limit.
$\lim_{x\rarr0}x^3+5x^2-7x+3$

• A. $0$
• B. $2$
• C. $3$
• D. $5$

Answer: (C)

Find the limit.
$\lim_{x\rarr2}\sqrt{x^2+5}$

• A. $0$
• B. $2$
• C. $3$
• D. $5$

Answer: (C)

Find the limit.
$\lim_{x\rarr3}\frac{x^2+2x-3}{x-2}$

• A. $0$
• B. $3$
• C. $12$
• D. $DNE$

Answer: (C)

Limits of Rational Functions: Denominator = 0

Understanding limits in rational functions is crucial for calculus. When evaluating limits, if the denominator does not equal zero at a specific point, you can directly substitute that value into the function. For example, consider the function:

\[\frac{x^2 + 3x + 2}{x + 1}\]

To find the limit as \(x\) approaches 0, simply substitute 0 into the function:

\[\frac{0^2 + 3(0) + 2}{0 + 1} = \frac{2}{1} = 2\]

However, if you want to find the limit as \(x\) approaches -1, you first check the denominator:

\[x + 1 = -1 + 1 = 0\]

Since the denominator equals zero, you cannot substitute directly. Instead, you should factor both the numerator and the denominator. The numerator \(x^2 + 3x + 2\) factors to:

\[ (x + 1)(x + 2) \]

Thus, the function can be rewritten as:

\[\frac{(x + 1)(x + 2)}{(x + 1)}\]

Now, you can cancel the common factor \(x + 1\), leading to:

\[x + 2\]

Now, you can evaluate the limit as \(x\) approaches -1:

\[-1 + 2 = 1\]

In summary, when faced with a rational function where the denominator equals zero, the steps are:

1. Check if the denominator equals zero at the limit point.
2. If it does, factor both the numerator and denominator.
3. Cancel any common factors.
4. Evaluate the limit using the simplified function.

For another example, consider:

\[\frac{x^2 + 2x - 15}{x - 3}\]

To find the limit as \(x\) approaches 3, check the denominator:

\[3 - 3 = 0\]

Next, factor the numerator, which factors to:

\[ (x - 3)(x + 5) \]

Now, the function becomes:

\[\frac{(x - 3)(x + 5)}{(x - 3)}\]

Cancel the common factor \(x - 3\) to get:

\[x + 5\]

Now, evaluate the limit as \(x\) approaches 3:

\[3 + 5 = 8\]

Lastly, consider:

\[\frac{x + 2}{x^2 - x - 6}\]

To find the limit as \(x\) approaches -2, check the denominator:

\[(-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0\]

Factor the denominator \(x^2 - x - 6\) to:

\[ (x + 2)(x - 3) \]

Now the function is:

\[\frac{x + 2}{(x + 2)(x - 3)}\]

Cancel the common factor \(x + 2\) to simplify to:

\[\frac{1}{x - 3}\]

Now, evaluate the limit as \(x\) approaches -2:

\[\frac{1}{-2 - 3} = \frac{1}{-5} = -\frac{1}{5}\]

By following these steps, you can effectively find limits for rational functions, even when the denominator approaches zero.

Limits of Rational Functions: Denominator = 0 Example 1

Understanding how to find the limit of a rational function is essential in calculus, especially when dealing with cases where the denominator approaches zero. Let's explore the process through two examples that illustrate different scenarios.

In the first example, we need to find the limit of the function \( \frac{x^3 - 2x^2 + x}{x - 1} \) as \( x \) approaches 1. Substituting \( x = 1 \) directly into the denominator results in \( 1 - 1 = 0 \), indicating that we must factor the function to resolve the indeterminate form.

To factor the numerator, we first notice that each term contains an \( x \), allowing us to factor out \( x \):

\( x(x^2 - 2x + 1) \).

Next, we recognize that \( x^2 - 2x + 1 \) is a perfect square, which can be factored further as \( (x - 1)^2 \). Thus, the entire expression becomes:

\( \frac{x(x - 1)^2}{x - 1} \).

We can simplify this by canceling one \( (x - 1) \) from the numerator and denominator, leading to:

\( x(x - 1) \).

Now, we can evaluate the limit as \( x \) approaches 1:

\( 1(1 - 1) = 1 \cdot 0 = 0 \).

Therefore, the limit of the function as \( x \) approaches 1 is 0.

In the second example, we find the limit of the same function as \( x \) approaches 2. Here, substituting \( x = 2 \) into the denominator gives us \( 2 - 1 = 1 \), which is not zero. This allows us to directly substitute \( x = 2 \) into the function:

\( \frac{2^3 - 2 \cdot 2^2 + 2}{2 - 1} \).

Calculating the numerator, we have:

\( 2^3 = 8 \), \( 2 \cdot 2^2 = 8 \), so \( 8 - 8 + 2 = 2 \).

Thus, the limit simplifies to:

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

In summary, when finding limits of rational functions, if the denominator approaches zero, factor the expression to cancel common terms before substituting. If the denominator does not approach zero, direct substitution is sufficient. This understanding is crucial for solving limits effectively in calculus.

Find the limit.
$\lim_{x\rarr0}\frac{3x^2+7x}{x}$

• A. $0$
• B. $7$
• C. $-3$
• D. DNE

Answer: (B)

Find the limit.
$\lim_{x\rarr2}\frac{x^2-7x+12}{x-3}$

• A. $-1$
• B. $-2$
• C. $0$
• D. DNE

Answer: (B)

Find the limit.
$\lim_{x\rarr-2}\frac{x^2-5x-14}{x+2}$

• A. $-9$
• B. $-2$
• C. $0$
• D. Does not exist

Answer: (A)

Find the limit.
$\lim_{x\rarr-\pi}\frac{\sin x}{x}$

• A. $-\frac{1}{\pi}$
• B. $0$
• C. $\pi$
• D. Does not exist

Answer: (B)

Limits of Rational Functions with Radicals

When working with rational functions, it's essential to understand that the limit of a function can often be evaluated directly unless the denominator equals zero. In cases where the denominator is zero, we can still find the limit by factoring and canceling common factors. However, when the rational function includes radicals, the process requires a different approach. Instead of factoring, we multiply both the numerator and denominator by the conjugate of the radical expression in the denominator.

For example, consider the limit of the function \(\frac{x - 2}{\sqrt{x} - \sqrt{2}}\) as \(x\) approaches 2. Direct substitution results in a denominator of zero, indicating that we need to manipulate the expression. We start by multiplying the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{x} + \sqrt{2}\). This results in:

\[\frac{(x - 2)(\sqrt{x} + \sqrt{2})}{(\sqrt{x} - \sqrt{2})(\sqrt{x} + \sqrt{2})}\]

Multiplying the conjugates in the denominator gives us \(x - 2\). This allows us to cancel the common factor \(x - 2\) from the numerator and denominator, simplifying our expression to \(\sqrt{x} + \sqrt{2}\). We can now evaluate the limit by substituting \(x = 2\), yielding:

\[\sqrt{2} + \sqrt{2} = 2\sqrt{2}\]

Thus, the limit of the function as \(x\) approaches 2 is \(2\sqrt{2}\).

In another example, we find the limit of \(\frac{\sqrt{x + 9} - 3}{x}\) as \(x\) approaches 0. Direct substitution results in a zero denominator, prompting us to multiply by the conjugate of the numerator, \(\sqrt{x + 9} + 3\). This gives us:

\[\frac{(\sqrt{x + 9} - 3)(\sqrt{x + 9} + 3)}{x(\sqrt{x + 9} + 3)}\]

Multiplying the conjugates in the numerator results in \(x + 9 - 9 = x\). The expression simplifies to:

\[\frac{x}{x(\sqrt{x + 9} + 3)}\]

We can cancel the \(x\) terms, leading to:

\[\frac{1}{\sqrt{x + 9} + 3}\]

Now, substituting \(x = 0\) gives us:

\[\frac{1}{\sqrt{0 + 9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}\]

Therefore, the limit of this rational function as \(x\) approaches 0 is \(\frac{1}{6}\).

In summary, when faced with rational functions that contain radicals and result in a zero denominator, the key steps involve multiplying by the conjugate, simplifying by canceling common factors, and then evaluating the limit. This method ensures that we can effectively find limits even in more complex scenarios.

Limits of Rational Functions with Radicals Example 3

To find the limit of the function \(\frac{\sqrt{4 - x} - 2}{x}\) as \(x\) approaches 0, we first recognize that direct substitution leads to an indeterminate form \(\frac{0}{0}\). In such cases, using the conjugate can simplify the expression. The conjugate of the numerator, \(\sqrt{4 - x} - 2\), is \(\sqrt{4 - x} + 2\).

We multiply both the numerator and the denominator by this conjugate:

\[\frac{(\sqrt{4 - x} - 2)(\sqrt{4 - x} + 2)}{x(\sqrt{4 - x} + 2)}\]

Expanding the numerator using the difference of squares gives:

\[(\sqrt{4 - x})^2 - 2^2 = (4 - x) - 4 = -x\]

Thus, the expression simplifies to:

\[\frac{-x}{x(\sqrt{4 - x} + 2)}\]

We can cancel \(x\) in the numerator and denominator (noting that \(x \neq 0\) in the limit process), leading to:

\[\frac{-1}{\sqrt{4 - x} + 2}\]

Now, we can safely substitute \(x = 0\) into the simplified expression:

\[\frac{-1}{\sqrt{4 - 0} + 2} = \frac{-1}{2 + 2} = \frac{-1}{4}\]

Therefore, the limit of the function as \(x\) approaches 0 is \(\frac{-1}{4}\).

Find the limit.
$\lim_{x\rarr3}\frac{\sqrt{x}-\sqrt3}{x-3}$

• A. $2\sqrt3$
• B. $0$
• C. $\frac{\sqrt3}{6}$
• D. Does not exist

Answer: (C)

Find the limit.
$\lim_{x\rarr0}\frac{\sqrt{x+1}-1}{x}$

• A. $0$
• B. $\frac12$
• C. $1$
• D. Does not exist

Answer: (B)

Find the limit.
$\lim_{x\rarr5}\frac{x-5}{\sqrt{x}-\sqrt5}$

• A. $\sqrt5$
• B. $2\sqrt5$
• C. $\frac{\sqrt5}{10}$
• D. Does not exist

Answer: (B)

Here’s what students ask on this topic:

How do you find the limit of a polynomial function as x approaches a specific value?

To find the limit of a polynomial function as x approaches a specific value, you can use direct substitution. This means you simply plug the value of x into the polynomial and evaluate it. For example, if you have the polynomial function f(x) = 6x³ + 3x² - x + 5 and you want to find the limit as x approaches 2, you substitute 2 for x: f(2) = 6(2)³ + 3(2)² - 2 + 5. This simplifies to 48 + 12 - 2 + 5 = 63. Therefore, the limit of the polynomial function as x approaches 2 is 63.

What should you do if the denominator of a rational function equals zero when finding a limit?

If the denominator of a rational function equals zero when finding a limit, you need to factor both the numerator and the denominator to cancel out any common factors. For example, consider the function f(x) = (x² + 3x + 2) / (x + 1) as x approaches -1. First, factor the numerator: (x + 1)(x + 2). The function becomes (x + 1)(x + 2) / (x + 1). Cancel the common factor (x + 1), leaving you with x + 2. Now, substitute -1 for x: f(-1) = -1 + 2 = 1. Therefore, the limit of the function as x approaches -1 is 1.

How do you find the limit of a rational function with radicals in the denominator?

To find the limit of a rational function with radicals in the denominator, multiply the numerator and the denominator by the conjugate of the denominator. For example, consider the function f(x) = (x - 2) / (√x - √2) as x approaches 2. The conjugate of the denominator is (√x + √2). Multiply both the numerator and the denominator by this conjugate: (x - 2)(√x + √2) / (√x - √2)(√x + √2). The denominator simplifies to x - 2, which cancels with the numerator, leaving √x + √2. Substitute 2 for x: f(2) = √2 + √2 = 2√2. Therefore, the limit of the function as x approaches 2 is 2√2.

What is the process for finding the limit of a rational function when the denominator is zero and contains radicals?

When finding the limit of a rational function where the denominator is zero and contains radicals, start by multiplying the numerator and the denominator by the conjugate of the term with radicals. For example, for the function f(x) = (√(x + 9) - 3) / x as x approaches 0, the conjugate of the numerator is (√(x + 9) + 3). Multiply both the numerator and the denominator by this conjugate: (√(x + 9) - 3)(√(x + 9) + 3) / x(√(x + 9) + 3). The numerator simplifies to x, which cancels with the denominator, leaving 1 / (√(x + 9) + 3). Substitute 0 for x: f(0) = 1 / (√9 + 3) = 1 / 6. Therefore, the limit of the function as x approaches 0 is 1/6.

How do you find the limit of a rational function when the denominator is zero and cannot be factored easily?

When the denominator of a rational function is zero and cannot be factored easily, use the method of multiplying by the conjugate. For example, for the function f(x) = (x - 2) / (√x - √2) as x approaches 2, multiply the numerator and the denominator by the conjugate of the denominator: (√x + √2). This gives (x - 2)(√x + √2) / (√x - √2)(√x + √2). The denominator simplifies to x - 2, which cancels with the numerator, leaving √x + √2. Substitute 2 for x: f(2) = √2 + √2 = 2√2. Therefore, the limit of the function as x approaches 2 is 2√2.