Problem 7 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}\) (A) exists and it equals \(\sqrt{2}\) (B) exists and it equals \(-\sqrt{2}\) (C) Does not exist because \((x-1) \rightarrow 0\) (D) Does not exist because left hand limit is not equal to right hand limit

Step-by-Step Solution

Answer

The limit exists and equals \( \sqrt{2} \) (Option A).

Step 1: Rewrite the Expression

The given limit is \( \lim_{x \to 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1} \). To simplify, we can use the trigonometric identity \( 1 - \cos \theta = 2\sin^2(\theta/2) \). Here, substitute \( \theta = 2(x-1) \), so the expression becomes \( \sqrt{2\sin^2((x-1))} \). Thus, rewrite the expression as \( \lim_{x \to 1} \frac{\sqrt{2}\sin(x-1)}{x-1} \).

Step 2: Apply L'Hôpital's Rule

The limit is in the indeterminate form \( \frac{0}{0} \) as \( x \to 1 \). Thus, apply L'Hôpital's Rule. Differentiate the numerator and the denominator. The derivative of \( \sqrt{2} \sin(x-1) \) is \( \sqrt{2} \cos(x-1) \) and the derivative of \( x-1 \) is \( 1 \).

Step 3: Evaluate the Limit

Substitute \( x = 1 \) into the differentiated expression, \( \sqrt{2} \cos(x-1) \), which gives \( \sqrt{2} \cos(0) = \sqrt{2} \cdot 1 = \sqrt{2} \). Thus, the limit evaluates to \( \sqrt{2} \).

Step 4: Conclusion

Since \( \lim_{x \to 1} \frac{\sqrt{2}\sin(x-1)}{x-1} = \sqrt{2} \), the limit exists and it equals \( \sqrt{2} \). Therefore, the correct answer is option (A).

Key Concepts

Trigonometric Limits

Trigonometric limits play a key role in calculus, especially when dealing with functions that include sine, cosine, and other trigonometric expressions. Such limits often involve evaluating the behavior of a trigonometric function as the input approaches a particular value.
For the given exercise, we can use a helpful identity:

• The identity for cosine, \( 1 - \cos(\theta) = 2\sin^2(\theta/2) \), simplifies the expression beautifully. By rewriting the provided function, it becomes easier to see how the values change as \( x \to 1 \).
• This step is crucial as it transforms the problem from a potentially complex trigonometric limit into a form that is easier to apply calculus rules to, like L'Hôpital's Rule.

Understanding these identities and their applications allows us to tackle problems involving trigonometric limits with more confidence and clarity. This knowledge is not only essential for solving such problems, but also for applying these principles to more complex calculus problems.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for evaluating limits, especially when faced with indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule allows us to differentiate the numerator and the denominator independently and then find the limit of the new quotient.
In the context of the exercise:

• We first identify the indeterminate form \( \frac{0}{0} \) as \( x \to 1 \).
• Applying L'Hôpital's Rule involves finding the derivatives: The derivative of \( \sqrt{2}\sin(x-1) \) leads to \( \sqrt{2}\cos(x-1) \), and the derivative of \( x-1 \) is just 1.
• The new expression \( \sqrt{2}\cos(x-1) \) can then be evaluated as \( x \to 1 \), simplifying our task to basic substitution.

L'Hôpital's Rule simplifies evaluating limits that initially appear undefined, turning a challenging problem into a straightforward calculation. Mastery of this rule greatly expands your problem-solving toolbox in calculus.

Indeterminate Forms

Indeterminate forms are expressions in calculus that do not initially lead to a definitively known solution. These expressions appear as limits where both the numerator and denominator simultaneously approach values leading to forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
In our exercise:

• The limit \( \frac{\sqrt{1-\cos 2(x-1)}}{x-1} \) as \( x \to 1 \) becomes \( \frac{0}{0} \), an indeterminate form.
• These forms require special techniques, like algebraic manipulation or L'Hôpital's Rule, to resolve.

Understanding and identifying indeterminate forms is crucial to solving limits effectively. By recognizing these forms, you can know when to apply techniques like L'Hôpital's Rule or seek alternative strategies such as simplifying trigonometric identities. This is vital in calculus, helping to bridge the gap between the apparent impossibility and practical solutions.