Problem 112 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow-1} \frac{x+1}{\sqrt[4]{x+17}-2}\\{\text { Ans. } 32\\} $$

Step-by-Step Solution

Answer

The correct answer is 0, as the limit of the given function as \(x\) approaches -1 is \(\frac{0}{12}\), which simplifies to 0.

Step 1: Checking if the function is in indeterminate form when x = -1

We will substitute \(x = -1\) into the given function to check if it is in indeterminate form. \(\frac{-1+1}{\sqrt[4]{-1+17}-2} = \frac{0}{\sqrt[4]{16}-2}\) The given function is not in the indeterminate form when \(x = -1\). However, we will still proceed with rationalization to find the limit.

Step 2: Rationalizing the expression

Rationalize the denominator by multiplying both the numerator and the denominator with the conjugate of the denominator, which is \((\sqrt[4]{x+17}+2)\). \(\lim _{x \rightarrow-1} \frac{(x+1)(\sqrt[4]{x+17}+2)}{(\sqrt[4]{x+17}-2)(\sqrt[4]{x+17}+2)}\) By difference of squares, the denominator simplifies to: \(\lim _{x \rightarrow-1} \frac{(x+1)(\sqrt[4]{x+17}+2)}{((x+17)-4)}\)

Step 3: Simplifying the expression

Now, we can simplify the expression further: \(\lim _{x \rightarrow-1} \frac{(x+1)(\sqrt[4]{x+17}+2)}{(x+13)}\)

Step 4: Evaluating the limit

Finally, substitute \(x = -1\) into the simplified expression to evaluate the limit: \(\frac{(-1+1)(\sqrt[4]{-1+17}+2)}{(-1+13)} = \frac{0}{12} = 0\) Therefore, the limit of the given function as \(x\) approaches -1 is 0, not 32 as mentioned in the problem statement.

Key Concepts

Rationalization

Rationalization is a technique used when dealing with limits involving radicals or roots. It often helps simplify expressions by eliminating the radical from the denominator.
To rationalize a denominator, we multiply both the numerator and the denominator by the conjugate of the denominator.

• Conjugate Utilization: For an expression like \(\sqrt[4]{x+17} - 2\), the conjugate is \(\sqrt[4]{x+17} + 2\).
• Multiplication: By multiplying the entire expression by this conjugate, we alter the form into something easier to manage without changing the expression's value.

This process allows us to simplify and solve limits more effectively.

Indeterminate Form

Indeterminate forms occur in calculus when an expression is undefined, like \(\frac{0}{0}\). These forms hint that simple substitution won't work, so we need alternative methods like L'Hôpital's Rule or rationalization. In our exercise:

• What We Saw: Substitution gave \(\frac{0}{2}\), not an indeterminate form.
• Why Continue: We still rationalize to simplify and ensure we are correctly evaluating the limit.

Understanding indeterminate forms is crucial for knowing when to apply special techniques.

Difference of Squares

The difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\) is a powerful algebra tool. It's used here to simplify the denominator during rationalization.
In our expression:

• Expression Applied: \((\sqrt[4]{x+17} - 2)(\sqrt[4]{x+17} + 2)\) becomes \((x+17) - 4\).
• Simplification: This simplifies to \((x + 13)\), which is easier to handle.

Recognizing these patterns helps in efficient simplification.

Evaluating Limits

After simplifying an expression, the final step is evaluating the limit by substitution. This approach is straightforward once the denominator and numerator have been rationalized or simplified.

• Substitution: Plug \(x = -1\) into \(\frac{(x+1)(\sqrt[4]{x+17}+2)}{x+13}\).
• Result: This produces \(\frac{0}{12}\), showing the limit as zero.

The process demonstrates how algebraic manipulation helps find limits even when direct substitution is not initially evident.