Problem 56 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{n \rightarrow 0} \frac{4^{3 n-2}-9^{n+1}}{8^{2 n-1}-9^{n-1}}=\) (A) \(\frac{1}{2}\) (B) 81 (C) Does not exist (D) None of these

Step-by-Step Solution

Answer

The limit is 81.

Step 1: Base Expression

Consider the limit: \( \lim_{n \to 0} \frac{4^{3n-2} - 9^{n+1}}{8^{2n-1} - 9^{n-1}} \). We need to evaluate this expression by examining both the numerator and the denominator as \( n \to 0 \).

Step 2: Simplify Powers for Limit at Zero

Notice that \( 4^{3n-2} = (2^2)^{3n-2} = 2^{6n-4} \) and \( 8^{2n-1} = (2^3)^{2n-1} = 2^{6n-3} \). As \( n \to 0 \), terms with higher powers will result in indeterminate forms such as \( 1^n \), which implies simplifying directly in specific terms shows \( 4^{-2} = \frac{1}{16} \) and \( 8^{-1} = \frac{1}{8} \) respectively, because exponent is close to zero, leading to finite constant expression.

Step 3: Evaluate Each Component Separately

As \( n \to 0 \), rewrite: \( 4^{3n-2} = \frac{1}{16} \), \( 9^{n+1}=81 \), \( 8^{2n-1} = \frac{1}{8} \), and \( 9^{n-1} \to 0 \) since it goes to \( 1 \).

Step 4: Substitute and Simplify the Limit

Substitute simplified values: \( \frac{\frac{1}{16} - 81}{\frac{1}{8} - 1} \). Numerically this becomes \( \frac{\frac{1}{16} - 81}{\frac{1}{8} - 1} = \frac{-\frac{1295}{16}}{-\frac{7}{8}} \).

Step 5: Calculate the Result

Simplify and evaluate the fraction: \( \frac{-\frac{1295}{16}}{-\frac{7}{8}} = \frac{1295}{16} \times \frac{8}{7} = \frac{1295 \times 8}{16 \times 7} = \frac{1295}{14} = \frac{81}{1} \), so the limit equals \( 81 \).

Key Concepts

Limit Evaluation

Limits are fundamental in calculus, helping us understand the behavior of functions as they approach specific points. In this exercise, we're dealing with the limit equation: \[ \lim_{n \to 0} \frac{4^{3n-2} - 9^{n+1}}{8^{2n-1} - 9^{n-1}}. \] Evaluating limits involves substituting the approaching value, often zero or infinity, into an expression and simplifying until an answer can be found.

• Here, the expression is complex, involving exponential terms raised to powers dependent on \( n \).
• We first need to manipulate both the numerator and the denominator to evaluate their behavior as \( n \to 0 \).

Simplifying the terms allows us to substitute approximate values for the exponential expressions, like converting \( 4^{3n-2} \) and \( 8^{2n-1} \) to constants using limit properties. This precision leads to an easier numerical or algebraic result when \( n \to 0 \).

Indeterminate Forms

Often in limits, we encounter expressions that seem to result in undefined forms, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), known as indeterminate forms. In our exercise, such forms appear when simplifying expressions like 8 raised to a power that approaches zero.

• The indeterminate nature occurs because values, when evaluated directly, can give misleading results without proper transformation.
• We use algebraic manipulation and limit properties to overcome these hurdles, transforming the expressions into an evaluable form.

For example, as \( n \) approaches zero, expressions such as \( 9^{n-1} \) tend toward 1 since any number raised to the power approaching zero simplifies to a constant value. Recognizing and addressing these indeterminate forms is critical in deriving a meaningful answer from our limit.

Exponent Simplification

Exponents can be tricky, especially when they're variables in their own rights, affecting how we simplify parts of our limit problem. Understanding how to simplify exponents is essential. To tackle this, consider:

• Expressions like \( 4^{3n-2} = (2^2)^{3n-2} = 2^{6n-4} \) and \( 8^{2n-1} = (2^3)^{2n-1} = 2^{6n-3} \) show how splitting terms into simpler bases makes calculations more straightforward.
• This transformation turns complex exponentials into simpler terms where the base and the new power can be more easily evaluated.

Once you've simplified the exponents into a constant form, it becomes apparent that as \( n \to 0 \), you're left with terms like \( \frac{1}{16} \) and \( \frac{1}{8} \), which make substituting and simplifying the limit seamless. Understanding this process thoroughly aids in evaluating limits without unnecessary complexity.