Problem 56 - Advanced Calculus

Question

Find a) \(\lim _{\mathrm{x} \rightarrow \mathrm{C}}\left[\left(\mathrm{x}_{\mathrm{n}}-\mathrm{C}_{\mathrm{n}}\right) /(\mathrm{x}-\mathrm{C})\right]\) b) \(\lim _{\mathrm{x} \rightarrow 0}\left[\left(\mathrm{x}^{2}\right) /\left\\{\left(\mathrm{x}^{2}+1\right)^{(1 / 2)}-1\right\\}\right] .\)

Step-by-Step Solution

Answer

In summary, for the given functions: a) The limit of \(\frac{x_n - C_n}{x - C}\) as \(x \rightarrow C\) cannot be directly evaluated. More information is needed, potentially with the application of L'Hopital's Rule. b) The limit of \(\frac{x^2}{\left\{ (x^2 + 1)^{(1/2)} - 1 \right\}}\) as \(x \rightarrow 0\) is equal to 2 after applying L'Hopital's Rule.

Step 1: Part a: Analyzing the function

The function is given by \(\frac{x_n - C_n}{x - C}\). We need to find the limit as \(x\) approaches \(C\). We can use the property of limits that states: if \(a_n\) approaches \(A\) as \(n\) goes to infinity, and \(b_n\) approaches \(B\) as \(n\) goes to infinity, then \(\frac{a_n}{b_n}\) will approach \(\frac{A}{B}\) as long as \(B\) is not equal to zero.

Step 2: Part a: Finding the limit

Now let's apply the property of limits to the given function: \(\lim_{x \rightarrow C}\left[\frac{x_n - C_n}{x - C}\right] = \frac{\lim_{x \rightarrow C}(x_n - C_n)}{\lim_{x \rightarrow C}(x - C)}\) As \(x\) approaches \(C\), both \(x_n - C_n\) and \(x - C\) approach 0, so the limit is: \(\frac{\lim_{x \rightarrow C}(x_n - C_n)}{\lim_{x \rightarrow C}(x - C)} = \frac{0}{0}\) Since this is an indeterminate form, we cannot find the limit directly. However, since the function has a ratio of difference between the sequences, it is likely that L'Hopital's Rule can be applied.

Step 3: Part b: Analyzing the function

The function is given by \(\frac{x^2}{\left\{ (x^2 + 1)^{(1/2)} - 1 \right\}}\). We need to find the limit as \(x\) approaches 0. This function seems more involved as it involves a square root and a difference in a denominator.

Step 4: Part b: Finding the limit

As we need to find the limit at \(x=0\), we plug it into the function to see if it's an indeterminate form: \(\lim_{x \rightarrow 0} \frac{x^2}{\left\{ (x^2 + 1)^{(1/2)} - 1 \right\}} = \frac{0}{\left\{ (0 + 1)^{(1/2)} - 1 \right\}} = \frac{0}{0}\) Since we have an indeterminate form again, we can try to apply L'Hopital's Rule, which can be applied when both the numerator and denominator tend to 0 or infinity, and we have a quotient of derivatives.

Step 5: Part b: Applying L'Hopital's Rule

To find the derivatives, we have: \(u(x) = x^2, u'(x) = 2x; v(x) = (x^2 + 1)^{(1/2)}, v'(x) = \frac{1}{2}(x^2 + 1)^{(-1/2)}(2x)\) Now, applying L'Hopital's Rule, we get: \(\lim_{x\to 0} \frac{u'(x)}{v'(x)} = \lim_{x \rightarrow 0} \frac{2x}{\frac{1}{2}(x^2 + 1)^{(-1/2)}(2x)}\) Notice that we can simplify the expression in the limit: \(\lim_{x \rightarrow 0} \frac{2x}{x(x^2 + 1)^{(-1/2)}} = \lim_{x \rightarrow 0} \frac{2}{(x^2 + 1)^{(-1/2)}}\) Now evaluate the limit: \(\lim_{x \rightarrow 0} \frac{2}{(x^2 + 1)^{(-1/2)}} = \frac{2}{(0 + 1)^{(-1/2)}} = \frac{2}{1} = 2\) So, the second limit is equal to 2. In conclusion, we found that: a) The limit of \(\frac{x_n - C_n}{x - C}\) as \(x\) approaches \(C\) cannot be directly evaluated, and more information is required, potentially applying L'Hopital's Rule. b) The limit of \(\frac{x^2}{\left\{ (x^2 + 1)^{(1/2)} - 1 \right\}}\) as \(x\) approaches 0 is equal to 2.

Key Concepts

Indeterminate Forms

Indeterminate forms arise in calculus when evaluating a limit yields an undefined expression like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These situations occur when both the numerator and the denominator of a rational expression approach 0 or infinity simultaneously as the variable approaches a particular value.

In the exercise, we encountered indeterminate forms in both parts (a) and (b).

• In part (a), the expression \(\frac{x_n - C_n}{x - C}\) becomes \(\frac{0}{0}\) as \(x\) approaches \(C\).
• Similarly, in part (b), the expression \(\frac{x^2}{\left\{ (x^2 + 1)^{(1/2)} - 1 \right\}}\) also results in a \(\frac{0}{0}\) form as \(x\) approaches 0.

When faced with an indeterminate form, direct substitution in the limit won't work, and further analysis using special techniques, like **L'Hopital's Rule**, is required to find the actual limit.

L'Hopital's Rule

L'Hopital's Rule is a powerful technique in calculus for evaluating limits that result in indeterminate forms. When both the numerator and denominator of a function approaching a limit tend toward 0 or \(\infty\), L'Hopital's Rule can help properly evaluate the limit by differentiating the numerator and denominator separately.

The rule can be applied when:

• The limit is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• The derivatives of the numerator and the denominator exist and are continuous around the point of interest.

In part (b) of the exercise, L'Hopital's Rule was utilized to resolve the indeterminate form \(\frac{0}{0}\) of \(\frac{x^2}{\left\{ (x^2 + 1)^{(1/2)} - 1 \right\}}\). After applying differentiation to both the numerator (\(x^2\)) and the denominator \((x^2 + 1)^{(1/2)} - 1\), the limit was simplified leading to a result of 2 as \(x\) tends to 0. This showcases how L'Hopital's Rule can make complex limit evaluations much more feasible.

Limit Evaluation Techniques

Limit evaluation techniques are essential strategies used to determine the limits of functions at specific points. Besides L'Hopital's Rule, there are multiple methods such as factoring, conjugation, and substitution that help analyze and solve limits effectively.

In the exercise, the indeterminate forms \(\frac{0}{0}\) suggested the potential need for L'Hopital's Rule. However, before resorting to this rule, it is often possible to resolve simpler indeterminate forms through:

• Factoring: Useful when expressions in both the numerator and denominator share common factors.
• Conjugation: Especially helpful in cases with square roots, as it rationalizes the denominator.
• Direct Substitution: In straightforward scenarios where simplification occurs post substitution.

These techniques, used alone or in combination, can often solve limits without further complexity. Each problem may require a unique approach tailored to the given function's form.