Problem 34 - Calculus Early Transcendentals

Question

Limits of composite functions Evaluate each limit and justify your answer. $$\lim _{x \rightarrow 0}\left(\frac{x}{\sqrt{16 x+1}-1}\right)^{1 / 3}$$

Step-by-Step Solution

Answer

Answer: The limit is \(\frac{1}{2}\).

Step 1: Rewrite the cube root

Rewrite the given expression as a power of 1/3. This will help later in the analysis: $$\lim _{x \rightarrow 0}\left(\frac{x}{\sqrt{16 x+1}-1}\right)^{1 / 3} = \lim _{x \rightarrow 0}\left(\frac{x}{\sqrt{16 x+1}-1}\right)^{\frac{1}{3}}$$

Step 2: Rationalize the denominator

To simplify the expression further, we will need to rationalize the denominator within the expression inside the limit. We can do this by multiplying the numerator and the denominator by the conjugate of the denominator: $$\lim _{x \rightarrow 0}\frac{x\left(\sqrt{16 x+1}+1\right)}{\left(\sqrt{16 x+1}-1\right)\left(\sqrt{16 x+1}+1\right)}$$ Now, multiply out the denominator using the difference of squares: $$=\lim _{x \rightarrow 0}\frac{x\left(\sqrt{16 x+1}+1\right)}{(16x+1)-1}$$

Step 3: Simplify the expression

Now, simplify the expression inside the limit further by cancelling out common terms: $$=\lim _{x \rightarrow 0}\frac{x\left(\sqrt{16 x+1}+1\right)}{16x}$$ Now, cancel out the common term x: $$=\lim _{x \rightarrow 0}\frac{\sqrt{16 x+1}+1}{16}$$

Step 4: Take the limit

Now that the expression is simplified, we can easily take the limit as \(x\) approaches 0: $$\begin{aligned} \lim _{x \rightarrow 0}\frac{\sqrt{16 x+1}+1}{16}&=\frac{\sqrt{16(0)+1}+1}{16}\\ & = \frac{\sqrt{1}+1}{16} \\ & = \frac{2}{16} \end{aligned}$$

Step 5: Apply the cube root

Finally, apply the cube root (power of 1/3) that we rewrote in step 1: $$\left(\frac{2}{16}\right)^{\frac{1}{3}} = \left(\frac{1}{8}\right)^{\frac{1}{3}} = \frac{1}{2}$$ So, the final result is: $$\lim _{x \rightarrow 0}\left(\frac{x}{\sqrt{16 x+1}-1}\right)^{1 / 3} = \frac{1}{2}$$

Key Concepts

Rationalization

Rationalization is a tool often used to simplify math expressions and particularly useful when you're dealing with limits involving roots. In this process, you eliminate square roots from the denominator of a fraction, making the expression easier to handle.

To rationalize the denominator, you multiply both the numerator and the denominator by the conjugate of the denominator. For example, if your denominator is \(\sqrt{16x + 1} - 1\), its conjugate would be \(\sqrt{16x + 1} + 1\).

• Multiply the numerator and denominator by the conjugate, resulting in eliminating the root in the denominator.
• The difference of squares formula \((a - b)(a + b) = a^2 - b^2\) is applied.

By doing this, you transform the denominator to a more manageable form, such as \( (16x + 1) - 1 \). Rationalization is key in evaluating limits properly, avoiding indeterminate forms.

Difference of Squares

The difference of squares is a specific mathematical identity that helps simplify expressions involving the subtraction of one square from another. The formula is \((a - b)(a + b) = a^2 - b^2\).

This identity is especially useful in the process of rationalization as it helps in removing the square roots when simplifying an expression's denominator. When applying the conjugate, as we did with \( (\sqrt{16x + 1} - 1)(\sqrt{16x + 1} + 1) \), it simplifies to \( (16x + 1) - 1 \).

• Difference of squares turns the product of conjugates into a simple, non-root expression.
• This simplification aids in the cancellation of terms (like \(x\) in our case).

This process is integral in making the subsequent steps of limit evaluation smoother and more straightforward.

Cube Root

Taking a cube root means finding a number which when multiplied by itself three times gives the original number. It's often represented as raising a number to the power of \(\frac{1}{3}\).

In limit problems with composite functions, handling cube roots is common. However, it's essential to simplify other parts of the expression before dealing with the cube root. After simplifying:

• Apply the cube root to the simplified limit expression.
• Raise the result under the radical to the power of \(\frac{1}{3}\).

The original limit problem transforms from a complex composite structure to an easier-to-manage expression, allowing us to determine the final result by following standard cube root rules.

Limit Evaluation

Limit evaluation is a fundamental concept in calculus used to find what value a function approaches as the input gets closer to a specific point. In this case, as \(x\) approaches zero.

After simplifying the original expression, the limit can be computed effortlessly. Steps to evaluate the limit include:

• Simplify the expression as much as possible, remove any indeterminate forms.
• Direct substitution if possible, once the expression is simplified.

For the given example: once it's simplified to \(\frac{\sqrt{16x+1}+1}{16}\), you substitute \(x = 0\), giving \(\frac{\sqrt{1}+1}{16} = \frac{2}{16}\).
Once the value is found, any remaining operations, like taking the cube root, provide us with the final limit result, \(\frac{1}{2}\). Thus, understanding each step and how limits work helps guide you through even complex functions.