Problem 64 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

\lim _{x \rightarrow 3} \frac{x^{2}-6 x+9}{2 x^{2}+x-21}\\{\text { Ans. } 0\\}

Step-by-Step Solution

Answer

The limit \(\lim_{x \rightarrow 3} \frac{x^2 - 6x + 9}{2x^2 + x - 21}\) can be evaluated by first factoring the expression as \(\frac{(x-3)^2}{(2x - 7)(x + 3)}\). Since the numerator becomes 0 at \(x=3\) and the denominator doesn't, the limit is 0.

Step 1: Plug in 3 into the function

First, let's try to plug in the value \(x = 3\) into the function. \(\frac{(3)^2 - 6(3) +9}{2(3)^2 + (3) -21} = \frac{9 - 18 + 9}{18 + 3 - 21}\) This results in an indeterminate form \(\frac{0}{0}\), so we'll need another approach to find the limit.

Step 2: Factorization of the numerator and denominator

To proceed, we'll try factoring the numerator and the denominator if possible. \(x^2 - 6x + 9 = (x - 3)^2\) \(2x^2 + x - 21 = (2x - 7)(x + 3)\) Now the re-write the function: \(\frac{(x-3)^2}{(2x - 7)(x + 3)}\)

Step 3: Analyze the limit with the simplified function

We can see that the numerator becomes zero when \(x\) approaches 3, however, the denominator does not become zero when \(x\) approaches 3. This means that the function approaches zero as \(x\) approaches 3. Therefore, \(\lim_{x \rightarrow 3} \frac{x^2 - 6x + 9}{2x^2 + x - 21} = 0\).

Key Concepts

Indeterminate Form

Indeterminate forms often appear when evaluating limits, and they indicate that direct substitution doesn't yield a clear answer. When you substitute a specific value into a function and get a form like \(\frac{0}{0}\), it is an indeterminate form. This means the limit requires further analysis.

• An indeterminate form doesn't define a limit directly, so simply substituting the variable value isn't enough.
• Common indeterminate forms include \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\).

Understanding these forms is crucial because they guide us to use other methods such as factoring, rationalizing, or L'Hôpital's rule to solve the limit problem. Factoring is often the first step as it simplifies expressions, helping to resolve the zeroes in indeterminate forms.

Factoring Polynomials

Factoring transforms an expression into a product of simpler expressions, known as factors. In the context of limits, factoring helps by simplifying the function, particularly when dealing with **indeterminate forms**.

• Factoring polynomials involves finding expressions whose products equal the original polynomial. It simplifies computation, particularly for limits.
• In the exercise, the polynomial \(x^2 - 6x + 9\) was factored into \((x-3)^2\).
• The denominator \(2x^2 + x - 21\) was factored into \((2x-7)(x+3)\).

By factoring, potential common terms in the numerator and denominator cancel out, allowing the limit to be evaluated without indeterminacy. This is a powerful technique when polynomial forms exhibit zeros at particular values of \(x\).

Simplifying Rational Functions

Rational functions, being ratios of two polynomials, can often be challenging due to complex expressions. Simplification makes them manageable, especially when evaluating limits.

• Simplification is needed to cancel terms that cause indeterminacy, like \(\frac{0}{0}\).
• Once a polynomial is factored, it might reveal common factors in the numerator and the denominator that can be cancelled.
• In the given exercise, the factor \((x-3)\) can potentially cancel out between the numerator and a zero in the denominator, but only if no zero remains in the denominator.

Once the rational function is simplified, the limit can be evaluated directly, avoiding the issue of indeterminate forms. This process reduces computation complexity and clarifies the behavior as \(x\) approaches a specific value, like 3 in the given example.