Question

\(\lim _{\mathrm{x} \rightarrow 0}\left[\left\\{(1+5 \mathrm{x})^{(3 / \mathrm{n})}-1\right\\} / \mathrm{x}\right]=? \quad \mathrm{n} \in \mathrm{R}^{+}\) (a) \((5 / \mathrm{n})\) (b) \((10 / \mathrm{n})\) (c) \((15 / \mathrm{n})\) (d) \((1 / 5 \mathrm{n})\)

Short answer

The short answer for the given question is: \(\frac{15}{n}\).

Step-by-step solution

Identify the indeterminate form

First, we will verify if the expression is in an indeterminate form as x approaches 0. Given expression: \(\lim _{x \rightarrow 0}\left[\frac{(1+5 x)^{\left(\frac{3}{n}\right)}-1}{x}\right]\) As x approaches 0, the expression becomes: \[\frac{(1+5(0))^{\left(\frac{3}{n}\right)}-1}{0} = \frac{1 - 1}{0} = \frac{0}{0}\] So, it is in the indeterminate form \(\frac{0}{0}\).

Apply L'Hopital's Rule

L'Hopital's rule can be applied when the limit converges to the indeterminate form \(\frac{0}{0}\). In this case, we can differentiate the numerator and the denominator separately with respect to x, and then find the limit. The given expression: \(\lim _{x \rightarrow 0}\left[\frac{(1+5 x)^{\left(\frac{3}{n}\right)}-1}{x}\right]\) Let's differentiate the numerator and the denominator with respect to x: Numerator: \(\frac{d}{dx}((1+5x)^{\frac{3}{n}}-1)\) Using the chain rule and simplifying, we get: \(\frac{15}{n}(1+5x)^{\frac{3}{n}-1}\) Denominator: \(\frac{d}{dx}(x)=1\) Then, apply L'Hopital's rule on the given expression: \[\lim _{x \rightarrow 0}\frac{\frac{15}{n}(1+5x)^{\frac{3}{n}-1}}{1}\]

Find the limit

Now, we can find the limit of the expression as x approaches 0: \[\lim _{x \rightarrow 0}\frac{15}{n}(1+5x)^{\frac{3}{n}-1} = \frac{15}{n}(1+5(0))^{\frac{3}{n}-1} = \frac{15}{n}(1)^{\frac{3}{n}-1} = \frac{15}{n}\] It matches the option (c). So, the limit of the given expression as x approaches 0 is \(\frac{15}{n}\).

Key concepts

Indeterminate Forms

Indeterminate forms, like \(\frac{0}{0}\), occur in limit problems when direct substitution results in a meaningless expression. These forms don't provide clear information about the limit's value. In the original exercise, substituting \(x=0\) in the given limit, \(\lim _{x \rightarrow 0}\left[\frac{(1+5 x)^{(\frac{3}{n})}-1}{x}\right]\), results in \(\frac{0}{0}\), which is an indeterminate form.

This indeterminate form signals us to use additional techniques, such as L'Hopital's rule, to evaluate the limit. Recognizing indeterminate forms is crucial because it tells us that we haven't yet found the answer, but rather, we need to apply a strategic mathematical approach to find it.

Limits

Limits help us understand the behavior of functions as they approach a certain point. In this exercise, we deal with the limit of a function as \(x\) approaches 0. The concept of limits is fundamental in calculus as it helps in defining derivatives, integrals, and the continuity of functions.

The original function, \((1+5x)^{(\frac{3}{n})}-1\) divided by \(x\), was first evaluated at \(x=0\) to check for indeterminate forms. Once identified, limits guide us in analyzing what value the function approaches without actually reaching that point. Understanding limits aids in simplifying complex expressions and assessing pointwise behavior.

Chain Rule

The chain rule is a method for computing the derivative of a composite function. In our step-by-step solution, it was used to differentiate the function \((1+5x)^{\frac{3}{n}}-1\). This consists of an outer function, \(u^k\), where \( u = (1+5x)\), and an inner function, \(1+5x\), that are nested.

Applying the chain rule involves first differentiating the outer function with respect to the inner function and then multiplying it by the derivative of the inner function. This is how we arrived at \(\frac{15}{n}(1+5x)^{\frac{3}{n}-1}\), enabling us to apply L'Hopital's rule. The chain rule is indispensable for handling complex differentiations involving functions within other functions.

Differentiation

Differentiation, in calculus, refers to the process of finding the derivative of a function, representing how it changes. In resolving the given indeterminate form, differentiation was key to applying L'Hopital's rule. The derivative of the numerator was taken using techniques like the chain rule.

For this problem, differentiating \((1+5x)^{\frac{3}{n}}-1\) enabled us to transform the expression into \(\frac{15}{n}(1+5x)^{\frac{3}{n}-1}\). This step is critical as it simplified the problem, allowing us to evaluate the limit effectively. Differentiation helps solve a wide array of problems concerning rates of change and slope analyses.