Question

\(\lim _{\mathrm{x} \rightarrow 1}[\sqrt{\\{1-\operatorname{Cos} 2(\mathrm{x}-1)\\} /(\mathrm{x}-1)}]=?\) (a) \(\sqrt{2}\) (b) 1 (c) Limit does not exist (d) \(-\sqrt{2}\)

Short answer

(c) Limit does not exist

Step-by-step solution

Identify the indeterminate form

First, let's evaluate the expression when \(x \rightarrow 1\): \(\frac{\sqrt{1-\operatorname{Cos}2(x-1)}}{x-1}\) As x approaches 1, both the numerator and denominator approach 0. Since 0/0 is an indeterminate form, we'll need to use L'Hopital's Rule in the next step.

Apply L'Hopital's Rule

L'Hopital's Rule states that if the limit of a function, as x approaches a, is of the indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then the limit of the function is equal to the limit of the ratio of their derivatives: \(\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}\) In this case, let \(f(x) = \sqrt{1-\operatorname{Cos}2(x-1)}\) and \(g(x) = x-1\). Let's find their derivatives.

Differentiate the functions

To differentiate the functions, we have: \(f'(x) = \frac{d}{dx} \sqrt{1-\operatorname{Cos}2(x-1)} = \frac{-\operatorname{Sin}2(x-1)(2)}{2\sqrt{1-\operatorname{Cos}2(x-1)}}\) \(g'(x) = \frac{d}{dx} (x-1) = 1\) Now, we'll apply L'Hopital's Rule.

Apply L'Hopital's Rule and evaluate the limit

Using L'Hopital's Rule, we now have: \(\lim_{x \rightarrow 1} \frac{f(x)}{g(x)} = \lim_{x \rightarrow 1} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 1} \frac{-\operatorname{Sin}2(x-1)(2)}{2\sqrt{1-\operatorname{Cos}2(x-1)}}\) Now, we'll evaluate the limit as \(x \rightarrow 1\): \(\lim_{x \rightarrow 1} \frac{-\operatorname{Sin}2(x-1)(2)}{2\sqrt{1-\operatorname{Cos}2(x-1)}} = \frac{-\operatorname{Sin}(0)(2)}{2\sqrt{1-\operatorname{Cos}(0)}}\) This simplifies to: \(\frac{-2(0)}{2\sqrt{1-1}} = \frac{0}{0}\) The limit is still of the indeterminate form. Let's apply L'Hopital's Rule once more.

Differentiate and apply L'Hopital's Rule again

Differentiate the numerator and the denominator of the new expression: \(f''(x) = \frac{d}{dx} \frac{-\operatorname{Sin}2(x-1)(2)}{2\sqrt{1-\operatorname{Cos}2(x-1)}} = \frac{-2\operatorname{Cos}2(x-1)(2)}{2(1-\operatorname{Cos}2(x-1))}\) \(g''(x) = \frac{d}{dx} 1 = 0\) Applying L'Hopital's Rule again: \(\lim_{x \rightarrow 1} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 1} \frac{f''(x)}{g''(x)} = \lim_{x \rightarrow 1} \frac{-2\operatorname{Cos}2(x-1)(2)}{2(1-\operatorname{Cos}2(x-1))}\) Now, we'll evaluate the limit as \(x \rightarrow 1\): \(\lim_{x \rightarrow 1} \frac{-2\operatorname{Cos}2(x-1)(2)}{2(1-\operatorname{Cos}2(x-1))} = \frac{-2\operatorname{Cos}(0)(2)}{2(1-\operatorname{Cos}(0))}\) This simplifies to: \(\frac{-2(1)(2)}{2(1-1)}\) Since the denominator is equal to 0, this leads us to conclude that the limit does not exist.

Final Answer

Therefore, the correct answer is: (c) Limit does not exist

Key concepts

Indeterminate Forms

In calculus, indeterminate forms often surface when we're evaluating limits. They are expressions that don't immediately convey a specific value and typically include forms like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0 \), \( \infty^0 \), and \( 1^\infty \). Let's take a closer look at \( \frac{0}{0} \) since it's pivotal in our exercise.

When you come across a \( \frac{0}{0} \) case, you're basically encountering a situation where both the numerator and the denominator of a fraction tend to zero as the variable approaches a certain value. This creates a hiccup, as simply plugging in the value won't give you the limit. To uncover the precise value, we might use L'Hopital's Rule, which effectively provides us with an alternative method to find the limit by examining the behavior of the function's derivatives.

Differentiation

Differentiation is a foundational concept in calculus, allowing us to find the derivative of a function. The derivative signifies the rate at which a function's value changes regarding the change in its input value. In practical terms, it gives us the slope of the function at any given point. The process involves taking an original function and applying differentiation rules to discover its derivative function.

In the context of L'Hopital's Rule and indeterminate forms, differentiation steps in to help us out. When we encounter a \( \frac{0}{0} \) indeterminate form, we differentiate the numerator and the denominator separately. After this, we often get a new function that may be easier to handle, or at least not indeterminate, allowing us to take the limit and resolve the original problem.

Limit of a Function

The concept of a limit is essential in the journey towards understanding continuous functions, derivatives, and integrals. A limit attempts to ascertain the value that a function approaches as the input (or the variable) approaches a specific value. When assessing limits, we sometimes find that directly substituting the variable's approaching value into the function doesn't work due to indeterminacy. That's where clever techniques like L'Hopital's Rule become indispensable.

In our exercise, we look for the limit as \( x \) approaches 1, but a straightforward substitution leads to an indeterminate form. By employing L'Hopital's Rule and differentiation, we seek to transform this stubborn indeterminacy into an understandable and computable limit. In cases where repeatedly applying L'Hopital's Rule still results in indeterminacy or an undefined expression, we may conclude that the limit does not exist, like with our exercise problem.