Problem 67 - Single Variable Calculus

Question

Find the limit, if it exists. \(\lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}(1-h)}{\sqrt{h}}\)

Step-by-Step Solution

Answer

Using L'Hôpital's rule and evaluating the limit, we find that the limit of the given function as h approaches 0 from the positive side is: \(\lim_{h\rightarrow 0^{+}}\frac{\cos^{-1}(1-h)}{\sqrt{h}} = 0\)

Step 1: Verify the indeterminate form 0/0

As h approaches 0 from the positive side: - The numerator, \(\cos ^{-1}(1-h)\), approaches \(\cos ^{-1}(1)\), which equals 0. - The denominator, \(\sqrt{h}\), also approaches 0. The expression is in the indeterminate form 0/0 as h approaches 0 from the positive side.

Step 2: Compute the derivatives of the numerator and the denominator

Now we must find the derivatives of the numerator and the denominator of the function. Numerator derivative: Let's denote \(u = 1 - h\). The derivative of \(\cos ^{-1}(u)\) is \(-\frac{1}{\sqrt{1-u^2}}\). So, using the chain rule, we have: \(\frac{d}{dh}[\cos^{-1}(1-h)] = -\frac{1}{\sqrt{1-(1-h)^2}}(-1) = \frac{1}{\sqrt{1-(1-h)^2}}\) Denominator derivative: The derivative of \(\sqrt{h}\) with respect to \(h\) is \(\frac{1}{2\sqrt{h}}\).

Step 3: Apply L'Hôpital's rule and evaluate the limit

Using L'Hôpital's rule, we have: \(\lim_{h\rightarrow 0^{+}}\frac{\cos^{-1}(1-h)}{\sqrt{h}} = \lim_{h\rightarrow 0^{+}}\frac{\frac{1}{\sqrt{1-(1-h)^2}}}{\frac{1}{2\sqrt{h}}}\) Simplify the expression: \(\lim_{h\rightarrow 0^{+}}\frac{2\sqrt{h}}{\sqrt{1-(1-h)^2}}\) Substitute the values as \(h\) approaches 0: \(\lim_{h\rightarrow 0^{+}}\frac{2\sqrt{0}}{\sqrt{1-(1-0)^2}} = \frac{0}{1}\)

Step 4: Final Answer

Therefore, the limit of the given function as h approaches 0 from the positive side is: \(\lim_{h\rightarrow 0^{+}}\frac{\cos^{-1}(1-h)}{\sqrt{h}} = 0\)

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a fantastic tool for tackling limits that result in indeterminate forms such as \(0/0\) or \(\infty/\infty\). When you have a limit that results in one of these indeterminate forms, L'Hôpital's Rule permits the substitution of the original limit with a limit of the derivatives of the numerator and the denominator.

Specifically, assuming that the functions \(f(x)\) and \(g(x)\) are differentiable, and their limits as \(x\) approaches a certain value result in an indeterminate form, L'Hôpital's Rule states that: \[\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\] provided the right-hand limit exists or leads to ±∞. However, if the new limit is still an indeterminate form, the rule may be applied repeatedly. It’s essential, though, to confirm that the conditions for applying the rule are met before using it.

Indeterminate Form

In calculus, an indeterminate form is a mathematical expression that involves undefined operations, such as \(0/0\) or \(\infty - \infty\). These forms do not have a clear, unique limit and thus require additional methods to evaluate them. The example above demonstrates the indeterminate form \(0/0\): both the numerator and the denominator are approaching zero as \(h\) approaches zero from the positive side.

Indeterminate forms are important in calculus because they often occur in the context of limits, as with L'Hôpital's Rule, which provides a technique for resolving the ambiguity in the limit. Understanding and identifying these forms is crucial for correctly applying advanced calculus tools to determine limits.

Derivative of Inverse Trigonometric Functions

Inverse trigonometric functions, like \(\cos^{-1}(x)\) or arcsin(x), have their own specific derivatives that are used when calculating rates of change or slopes of tangent lines on their curves. For example, the derivative of \(\cos^{-1}(x)\) is \(-\frac{1}{\sqrt{1-x^2}}\), assuming the value of \(x\) lies in the range of the function where it is defined.

This particular derivative is useful when inverse trigonometric functions appear in the numerator or denominator of a fraction where the limit is sought. In the exercise at hand, this derivative is a key step in using L'Hôpital's Rule as we seek to resolve the indeterminate form by differentiating both the numerator and the denominator.

Applying Chain Rule

The Chain Rule is an essential differentiation technique in calculus used for finding the derivative of composite functions. In essence, the Chain Rule states that the derivative of a composed function \(f(g(x))\) is the derivative of \(f\) with respect to \(g\) multiplied by the derivative of \(g\) with respect to \(x\): \(\frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}\).

This rule comes into play when you're working with functions nested within each other, as was seen in the exercise where the inverse cosine function had an inner function \(1-h\). To find the derivative of such a composition, the Chain Rule is applied to ‘unpack’ the nested functions step by step. This process is essential for making progress when evaluating the limit using L'Hôpital's Rule as it helps us to differentiate correctly to find a meaningful limit.