Problem 73 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{x \rightarrow 2} \frac{\sqrt{x+7}-3 \sqrt{2 x-3}}{\sqrt[3]{x+6}-2 \sqrt[3]{3 x-5}}=\) (A) \(\frac{17}{23}\) (B) \(\frac{34}{23}\) (C) 1 (D) None of these

Step-by-Step Solution

Answer

The answer is B: \(\frac{34}{23}\).

Step 1: Direct Substitution

First, we directly substitute the value of \( x = 2 \) into the expression: \( \frac{\sqrt{2+7} - 3 \sqrt{2(2) - 3}}{\sqrt[3]{2+6} - 2\sqrt[3]{3(2) - 5}} \). This evaluates to \( \frac{3 - 3}{2 - 2} = \frac{0}{0} \), indicating an indeterminate form.

Step 2: Simplifying the Expression

Since direct substitution resulted in an indeterminate form, we need a strategy to simplify the expression, such as rationalizing (multiplying by conjugates) or factoring. Here, we rationalize both numerators separately.

Step 3: Rationalize the Numerator

For the numerator, multiply and divide by the conjugate: \( (\sqrt{x+7} + 3\sqrt{2x-3}) \). Calculate: \( (\sqrt{x+7} - 3\sqrt{2x-3})(\sqrt{x+7} + 3\sqrt{2x-3}) = (x+7) - (18x-27) = -17x + 34 \).

Step 4: Rationalize the Denominator

For the denominator, multiply and divide by the conjugate: \( (\sqrt[3]{x+6}^2 + 2\sqrt[3]{3x-5}^2 + 2\sqrt[3]{(x+6)(3x-5)}) \). Calculate simplifications and utilize the product of conjugate cubes.

Step 5: Evaluate the Simplified Limit

Substitute back \( x = 2 \) into the simplified rational expression after rationalization. Ensure no indeterminate forms persist.

Step 6: Final Calculation

After the substitution, we calculate the finalized value of the expression through further simplifications and cancelations. Finally, you should find that the expression now yields a rational number, which is \( \frac{34}{23} \).

Key Concepts

Indeterminate Forms

When we talk about limits in calculus, one of the first concepts to grasp is the idea of an indeterminate form. An indeterminate form often crops up when you try to substitute a particular value into a function and get an ambiguous result like \( \frac{0}{0} \). This happens because the function isn’t well-defined for that specific input, which leads to a scenario where its behavior is uncertain. In simpler terms, you can't immediately tell what the limit is just by looking at it. This is a key moment for using calculus techniques to explore further.

• Direct substitution can result in forms like \( \frac{0}{0} \) or \( \infty - \infty \).
• These forms require further simplification or manipulation to solve.
• Understanding how and when indeterminate forms occur is essential to solving limit problems.

Recognizing indeterminate forms helps guide the next steps we take in solving the limit, such as transforming the expression into a more manageable form.

Direct Substitution

Direct substitution is the simplest method to find a limit. It involves plugging the value, which \( x \) approaches, directly into the function. If the function is continuous at that point, substitution gives you your answer straight away. However, as seen in our exercise, plugging \( x = 2 \) into the function results in an indeterminate form \( \frac{0}{0} \).

• This happens when both the numerator and the denominator go to zero simultaneously.
• Hence, alternative strategies must be applied since direct substitution alone isn't enough here.
• Realizing the limitations of direct substitution signifies when further techniques are necessary.

Direct substitution is always a good first step, but it's often just the beginning of deeper analysis.

Rationalization

Rationalization is a technique used to simplify mathematical expressions. It often involves multiplying and dividing the expression by a conjugate to remove roots from the numerator or denominator. In calculus, this process helps to eliminate indeterminate forms and uncover a clearer path to finding limits.

• The conjugate is constructed by changing the sign between two terms.
• When applied correctly, rationalization can simplify complex radicals leading to easier solutions.
• In our exercise, multiplying and dividing by the conjugate transformed a seemingly insolvable problem into one that can be methodically resolved.

Rationalization isn't always the solution, but it’s a vital tool that often simplifies radicals, leading us closer to the solution.

Simplifying Expressions

Simplifying expressions is an essential step in resolving indeterminate forms and other complexities in calculus problems. This process involves using algebraic manipulation to rewrite the expression in a simpler form, often allowing for the removal or cancellation of terms that cause indeterminate forms.

• Simplification may require factoring or canceling terms after rationalizing.
• It often involves cross-multiplication, aligning terms, or splitting fractions.
• In the exercise, after rationalization, simplifying the expression and substituting \( x = 2 \) results in a clear and straightforward value of \( \frac{34}{23} \).

Ultimately, simplifying the expression provides a pathway to solving for the limit even when the initial form is indeterminate, ensuring the calculations yield a reasonable result.