Question

\(\operatorname{Lt}_{x \rightarrow 0} \frac{\sqrt[5]{x+32}-2}{x}=\) (1) \(\frac{1}{16}\) (2) \(\frac{1}{45}\) (3) \(\frac{1}{80}\) (4) \(\frac{1}{100}\)

Short answer

Answer: The value of the limit is \(\boxed{\frac{1}{80}}\).

Step-by-step solution

Express the problem as a limit problem

We are given \(\operatorname{Lt}_{x \rightarrow 0} \frac{\sqrt[5]{x+32}-2}{x}\), which represents the limit of the function \(\frac{\sqrt[5]{x+32}-2}{x}\) as x approaches 0.

Rationalize the numerator

To rationalize the numerator, we write the difference of cubes as a product of two conjugates. We multiply both the numerator and denominator by the conjugate of the numerator, which is \((\sqrt[5]{x+32})^4 + (\sqrt[5]{x+32})^3⋅2 + (\sqrt[5]{x+32})^2⋅2^2 + (\sqrt[5]{x+32})⋅2^3 + 2^4\): $$\operatorname{Lt}_{x \rightarrow 0} \frac{(x+32) - 2^5}{x((\sqrt[5]{x+32})^4 + (\sqrt[5]{x+32})^3⋅2 + (\sqrt[5]{x+32})^2⋅2^2 + (\sqrt[5]{x+32})⋅2^3 + 2^4)}$$

Simplify the numerator

The numerator simplifies to the following expression: $$\operatorname{Lt}_{x \rightarrow 0} \frac{x-32}{x((\sqrt[5]{x+32})^4 + (\sqrt[5]{x+32})^3⋅2 + (\sqrt[5]{x+32})^2⋅2^2 + (\sqrt[5]{x+32})⋅2^3 + 2^4)}$$

Examine the behavior of the function as x approaches 0

As x approaches 0, we can replace x in the numerator with 0, since it simplifies the expression. The denominator, however, still contains x as a factor or in terms, so we need to analyze them separately. We can do this dividing both the numerator and denominator by x as follows: $$\operatorname{Lt}_{x \rightarrow 0} \frac{1-32/x}{((\sqrt[5]{x+32})^4 + (\sqrt[5]{x+32})^3⋅2 + (\sqrt[5]{x+32})^2⋅2^2 + (\sqrt[5]{x+32})⋅2^3 + 2^4)}$$

Evaluate the limit

Now, apply the limit to all terms in the expression as x approaches 0. We'll see that some terms go to 0, and others go to their finite values: $$\frac{1}{(2^4 + 2^3⋅2 + 2^2⋅2^2 + 2^3 + 2^4)} = \frac{1}{80}$$ The result is \(\frac{1}{80}\), which corresponds to option (3), so the correct answer is \(\boxed{\frac{1}{80}}\).

Key concepts

Rationalizing the Numerator

Rationalizing the numerator in limit problems is a technique used to simplify a limit expression so it's easier to evaluate. When dealing with square roots or any roots in the numerator, this process can make a limit more tractable. Essentially, the goal is to eliminate the roots in the numerator by multiplying the original expression by a form of 1 that contains the conjugate of the numerator's radical part. This gives us a difference of squares, cubes, or higher powers which can often help clear out indeterminate forms such as 0/0.

For instance, in our given problem, the numerator contains a fifth root, which poses a challenge as the value of x approaches 0. To overcome this, we multiply the numerator and the denominator by the conjugate of the numerator, effectively getting rid of the fifth root. This step is crucial, as it leads to a polynomial in the numerator that we can then analyze without the complications of the root.

Evaluating Limits

Evaluating limits is fundamental in calculus, as it allows us to determine the behavior of a function as it approaches a certain point. In the context of our problem, we're interested in the limit of the function as x approaches 0. Once the expression is simplified, possibly through rationalizing the numerator or other algebraic manipulations, the next step is to substitute the approaching value into the simplified expression wherever possible.

If direct substitution yields an indeterminate form, such as 0/0, further manipulation or techniques like L'Hôpital's rule might be needed. In our example, after rationalization and simplifying the numerator, we ended up with terms in x that we could then evaluate as x approaches 0. As a result, some terms vanish, while others assume finite values, enabling us to find the final value of the limit.

Limit of a Function

Understanding the concept of a limit of a function is vital for grasping the fundamentals of calculus. The limit describes the behavior of a function as the input (typically denoted as x) approaches a particular value. Limits can tell us the value that a function is getting closer to, which might not necessarily be the function's actual value at that point.

In the context of our problem, we have a function with an input variable x that we're considering as it gets closer to 0. The core idea is to see what value the function 'settles on' or approaches as we get arbitrarily close to x = 0. The ability to determine this approaching value, even when it cannot be calculated by simple substitution due to an indeterminate form, is what makes the concept of limits such a powerful tool in the study of calculus.