Question

\(\lim _{\mathrm{x} \rightarrow 1}\left[\\{\sin [\mathrm{n}(\sqrt{\mathrm{x}}-1)]\\} /\left({ }^{4} \sqrt{\mathrm{x}}-1\right)\right]=\ldots \ldots .\left(\mathrm{n} \in \mathrm{R}^{+}, \mathrm{x}>0\right)\) (a) \((\mathrm{n} / 2)\) (b) \(2 \mathrm{n}\) (c) \(n\) (d) \((\mathrm{n} / 4)\)

Short answer

\( \lim _{x \rightarrow 1}\left[\frac{\sin{[n(\sqrt{x}-1)]}}{4\sqrt{x}-1}\right]=\frac{n}{4} \)

Step-by-step solution

Identify the indeterminate form

As x approaches 1, let's check the behavior of the function's components: sin[n(√x -1)] and (4√x - 1).

Check the sine function term

When x approaches 1, we have √x - 1 approaching 0, therefore, the sine function will have n(√x -1) inside it approaching 0 as well. As a result, the sine function value will also approach 0.

Check the root term

When x approaches 1, we have (4√x - 1) approaching (4 - 1) which equals 3.

Determine the indeterminate form

Considering the behaviors of the two terms, the limit is in the indeterminate form \(0/3\). We can resolve this indeterminate form using L'Hôpital's Rule.

Apply L'Hôpital's Rule

To apply L'Hôpital's Rule, we need to differentiate both the numerator and denominator of the limit function with respect to x. So, we calculate: - The derivative of sin[n(√x -1)] with respect to x: \[\frac{d}{dx}(\sin[n(\sqrt{x} - 1)]) = \cos[n(\sqrt{x} - 1)] \cdot n \cdot \frac{1}{2\sqrt{x}}\] - The derivative of (4√x - 1) with respect to x: \[\frac{d}{dx}(4\sqrt{x} - 1) = 2\]

Calculate the new limit

Now that we have the derivatives calculated, we use L'Hôpital's Rule to calculate the limit: \[\lim_{x \to 1} \frac{\sin[n(\sqrt{x} - 1)]}{4\sqrt{x} - 1} = \lim_{x \rightarrow 1} \frac{\cos[n(\sqrt{x} - 1)] \cdot n \cdot \frac{1}{2\sqrt{x}}}{2}\] We can simplify this expression: \[\lim_{x \rightarrow 1} \frac{\cos[n(\sqrt{x} - 1)] \cdot \frac{n}{2\sqrt{x}}}{2}\] Now, we can substitute x = 1 to find the limit: \[\frac{\cos[n(\sqrt{1} - 1)] \cdot \frac{n}{2\sqrt{1}}}{2} = \frac{\cos[n(0)] \cdot \frac{n}{2}}{2}\] Since cos(0) = 1, we end up with: \[\frac{\frac{n}{2}}{2} = \frac{n}{4}\] Therefore, the answer is (d) n/4.

Key concepts

L'Hôpital's Rule

Understanding L'Hôpital's Rule can be a game-changer when dealing with complex calculus problems. It is a strategic tool used specifically to evaluate limits that present indeterminate forms like \(0/0\) or \(\infty/\infty\). The concept is named after the French mathematician Guillaume de l'Hôpital, who published the rule in his book.

L'Hôpital's Rule states that if we have an indeterminate form of type \(0/0\) or \(\infty/\infty\), then the limit of a ratio of two functions as they approach a point can be found by taking the limit of the ratio of their derivatives instead. This only works if certain conditions are met, primarily that the original functions must be differentiable near the point of interest and their derivatives should not result in an indeterminate form. Here's the role L'Hôpital's Rule plays:

• It simplifies the process of finding limits that otherwise would be cumbersome to compute.
• It provides a methodical approach that can be applied to many calculus problems.
• It encourages understanding the behaviour of functions through their derivatives.

In our textbook example, L'Hôpital's Rule is applied because the limit as x approaches 1 initially gives us an indeterminate form \(0/3\). By differentiating the numerator and the denominator separately and then taking their limit, we find the solution, reinforcing the rule’s utility in calculus.

Indeterminate Forms

Indeterminate forms pose significant challenges in calculus, especially when calculating limits. An indeterminate form is the uncertain form that a limit might seem to approach based on direct substitution. Common indeterminate forms include \(0/0\), \(\infty/\infty\), \(0\cdot\infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), and \(\infty^0\). These expressions do not yield a clear limit and require additional techniques to solve.

In practice, indeterminate forms indicate that more information is needed to determine the limit's value. Analyzing how the functions behave around the point of interest is essential. For instance, the exercise given presented us with the indeterminate form \(0/3\) at first glance. However, with closer inspection using calculus techniques such as L'Hôpital's Rule, we can resolve these forms and find the precise limits. Understanding indeterminate forms is not just about recognizing them but also about knowing how to approach and resolve them, thus enhancing our grasp of function behaviour.

Derivatives

Derivatives are fundamental to calculus, capturing the essence of change and enabling the measuring of how one quantity varies with respect to another. In simplest terms, the derivative of a function at a point is the slope of the tangent line to the function's graph at that point. Technically, it is the limit of the average rate of change of the function over an interval as the interval becomes infinitesimally small.

When we compute a derivative, we're finding a function that gives us the rate of change of one variable with respect to another. This concept is pivotal when applying L'Hôpital's Rule, as we replace the original functions with their derivatives to resolve indeterminate forms. Derivatives not only help us in solving limits but also have numerous applications in various fields such as physics, economics, and engineering where they represent rates like velocity, cost, and efficiency respectively.

• They reflect instantaneous rates of change, mirroring actual phenomena.
• Derivatives are used for finding extrema and solving optimization problems.
• They serve as the foundation for more advanced calculus concepts like integration.

In our textbook problem, derivatives transformed an indeterminate form into a solvable limit, showcasing the derivative's power as an analytical tool in calculus.