Problem 217 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \left.\lim _{x \rightarrow 1} \frac{\sin (1-x)}{\sqrt{x}-1} \text { \\{Ans. }-2\right\\} $$

Step-by-Step Solution

Answer

Using L'Hôpital's rule, we find the limit of the given function as follows: First, find the derivatives of the numerator and denominator: \(f'(x) = -\cos(1-x)\), \(g'(x) = \frac{1}{2x^{\frac{-1}{2}}}\) Next, apply L'Hôpital's rule to the limit: \(\lim_{x \rightarrow 1} \frac{-\cos(1-x)}{\frac{1}{2x^{\frac{-1}{2}}}}\) Using direct substitution, we get: \(\frac{-\cos(0)}{\frac{1}{2}}\) Finally, evaluating the simplified expression, we obtain the answer: \(\lim_{x \rightarrow 1} \frac{\sin(1-x)}{\sqrt{x}-1} = -2\)

Step 1: Recognize the indeterminate form

First, let's attempt direct substitution to see what form we are dealing with. As x approaches 1, we get: \[ \frac{\sin(1-1)}{\sqrt{1}-1} = \frac{\sin(0)}{0} \] This is an indeterminate form (0/0). Therefore, we need to use the L'Hôpital's rule or some other technique to manipulate the expression before finding the limit.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's rule states that if the limit of the ratio of the derivatives of two functions is finite, then the limit of their ratio is equal to that value as well. This rule is applied when facing an indeterminate form (0/0) or (∞/∞). In our case, we will find the derivatives of the numerator and denominator function: Numerator: \[ f(x) = \sin(1-x) \] Derivative of the numerator: \[ f'(x) = \cos(1-x)(-1) = -\cos(1-x) \] Denominator: \[ g(x) = \sqrt{x}-1 \] Derivative of the denominator: \[ g'(x) = \frac{d}{dx}(x^{\frac{1}{2}}) \] \[ g'(x) = \frac{1}{2x^{\frac{-1}{2}}} \]

Step 3: Calculate the new limit

Now, apply L'Hôpital's rule to the limit: \[ \lim_{x \rightarrow 1} \frac{-\cos(1-x)}{\frac{1}{2x^{\frac{-1}{2}}}} \] We can now attempt direct substitution again: \[ \frac{-\cos(1-1)}{\frac{1}{2(1)^{\frac{-1}{2}}}} \] Which simplifies to: \[ \frac{-\cos(0)}{\frac{1}{2}} \]

Step 4: Evaluate the simplified expression

Finally, evaluate the simplified expression: \[ \frac{-1}{\frac{1}{2}} = -2 \] So, the limit of the given function is -2: \[ \lim_{x \rightarrow 1} \frac{\sin(1-x)}{\sqrt{x}-1} = \boxed{-2} \]

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to evaluate limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). To apply this rule, we first verify that our limit is indeed in an indeterminate form. Once confirmed, we find the derivatives of the numerator and denominator independently and then take the limit of their quotient.

L'Hôpital's Rule can be summarized with the steps:

• Identify the indeterminate form.
• Differentiate the numerator and the denominator separately.
• Apply the limit to the new quotient.
• Repeat if the result is still an indeterminate form.

For the given problem, you identified \( \frac{\sin(0)}{0} \) as an indeterminate form \( \frac{0}{0} \), allowing the use of L'Hôpital's Rule. Derivatives were computed, and the new limit yielded a manageable expression, ultimately leading to the result \(-2\).

Indeterminate Forms

In calculus, an indeterminate form occurs when the limit of a function cannot be directly determined from its algebraic form. These forms typically arise as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) when performing limit evaluations.

Here’s why these forms are indeterminate:

• \( \frac{0}{0} \): The zero in the denominator indicates division by zero, which doesn't have a defined value. Yet, by manipulating the expression, we can often find a meaningful limit.
• \( \frac{\infty}{\infty} \): While technically undefined, it suggests that with some algebraic rearrangement or calculus techniques, we can possibly extract a finite limit.

The exercise initially involves an indeterminate form \( \frac{0}{0} \), revealing the need for a method like L'Hôpital's Rule to progress towards calculating a defined limit.

Trigonometric Limits

Trigonometric limits involve functions like sine, cosine, and tangent as variables approach certain values. These often appear in indeterminate forms that require special techniques like algebraic manipulation or the application of L'Hôpital's Rule.

In the problem, you have the sine function: \( \sin(1-x) \). As \( x \) approaches 1, \( 1-x \) approaches 0, making \( \sin(0) \). Understanding the behavior of sine in such situations helps simplify the approach to solving the limit.

The sine function has a known limit property: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Though not directly applied here, recognizing trigonometric limits is essential when resolving similar indeterminate forms where trigonometric identities take center stage.

Derivative Calculations

Derivative calculations are crucial when applying L'Hôpital's Rule since we need the derivatives of both the numerator and denominator of a function's quotient. Derivatives give us rates of change and are foundational in calculus for finding slopes of tangent lines and, in this context, for simplifying limits.

In the solved example, the derivatives were:

• For the numerator \( f(x) = \sin(1-x) \), the derivative is \( f'(x) = -\cos(1-x) \).
• For the denominator \( g(x) = \sqrt{x} - 1 \), the derivative is \( g'(x) = \frac{1}{2x^{\frac{-1}{2}}} \).

Calculating these derivatives reshaped the original problematic expression into a form where the limit could be applied directly, leading to a conclusive result. Derivatives, therefore, transition an indeterminate expression into a quantifiable outcome.