Problem 11 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The value of \(\lim _{h \rightarrow 0} \frac{\ln (1+2 h)-2 \ln (1+h)}{h^{2}}\) is (A) 1 (B) \(-1\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the Expression

We need to evaluate the limit \( \lim_{h \to 0} \frac{\ln(1+2h) - 2\ln(1+h)}{h^2} \). This involves a difference of logarithms in the numerator, which we may need to simplify before applying limit techniques.

Step 2: Use Logarithm Properties

Apply the property of logarithms that \( a\ln(b) = \ln(b^a) \). Rewrite the expression as \( \ln(1+2h) - \ln((1+h)^2) \). This simplifies to \( \ln\left(\frac{1+2h}{(1+h)^2}\right) \).

Step 3: Simplify the Logarithmic Expression

The expression inside the logarithm becomes \( \frac{1+2h}{1+2h+h^2} \). Further simplify it to \( \ln\left(\frac{1+2h}{1+2h+h^2}\right) = \ln(\frac{1+2h}{(1+h)^2}) = \ln(\frac{1}{1+h^2}) \).

Step 4: Apply Series Expansion

Expand the function \( \ln(1+x) \) around \( x = 0 \) using \( \ln(1+x) \approx x - \frac{x^2}{2} \) for small \( x \). Apply this to \( \ln(\frac{1+2h}{1+2h+h^2}) \).

Step 5: Substitute and Simplify the Limit

Substitute the expansion back into the limit expression to get an expression in \( h\) terms. After simplification, you'll find that dominant terms involving \( h \) will cancel out due to the \( h^2 \) factor in the denominator.

Step 6: Evaluate the Limit

Finally, evaluate the simplified expression as \( h \to 0 \) by canceling terms in the numerator proportional to \( h^3 \) and solving for the limit, which results in \( 0 \).

Key Concepts

Logarithm Properties

When tackling problems involving logarithms, understanding their properties can greatly simplify the expression you're working with. Here are a few key properties to keep in mind:

• The logarithm of a power: A logarithm, such as the expression in our original exercise, may include terms such as \( a\ln(b) \). This can be rewritten using the property \( a\ln(b) = \ln(b^a) \), which is particularly useful for condensation or simplification.
• The difference of logarithms can be expressed as a single logarithm: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). This is especially helpful when comparing or simplifying expressions in a single logarithm, making the algebra more manageable.

In our original problem, we utilized these properties to simplify the expression \( \ln(1+2h) - 2\ln(1+h) \) to \( \ln\left(\frac{1+2h}{(1+h)^2}\right) \), transforming the expression via simplification from differences into a more simplified single logarithmic term.
Remember, simplifying logarithmic expressions early on can save time and make the path to the solution much more straightforward.

Series Expansion

Series expansion is a powerful tool in calculus, particularly when dealing with limits and evaluating functions at points close to zero. A common series expansion used is the Taylor series expansion around zero, known as the Maclaurin series for functions.
For logarithms, particularly \( \ln(1+x) \), you can approximate it for small values of \( x \) using \[\ln(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\]This expression allows us to expand complex logarithmic functions into simpler polynomials.

• In our example, we used this pattern to expand \( \ln\left(\frac{1+2h}{1+2h+h^2}\right) \).
• Such expansion turns potentially complex limits into polynomials that are easier to manipulate and evaluate.

By substituting this expansion into the expression involving the limit, we were able to transform the limit problem into a polynomial form. This facilitated the simplification process and eventually led to canceling out terms that produced a straightforward evaluation as \( h \rightarrow 0 \).
Remember, series expansions are particularly helpful when you need to simplify expressions where direct computation would be challenging or infeasible.

Simplifying Expressions

Simplifying expressions in calculus, particularly those involving limits, is crucial for managing and solving complex mathematical problems. Simplification can involve several techniques, including algebraic manipulation and the use of known mathematical identities.

• In our exercise, we had to simplify an initial complex expression involving a difference of logarithms. This involved using logarithmic identities to combine the fractions spread across the numerator into a single logarithmic term.
• Once we achieved simplification to a single logarithmic expression, we employed the series expansion tool to translate our logarithmic component into a polynomial form by expanding it into its series counterpart, allowing easier calculation.

The primary benefit of simplification is that it reduces potential errors and makes subsequent steps in the calculation more direct and intuitive. Especially when the terms of one type (like \( h^2 \) in the denominator) dominate the behavior of the function as \( h \to 0 \), reducing expressions early on generally leads to faster and more accurate solutions.
Overall, effective simplification can transform intimidating expressions into something more manageable, enhancing your ability to find and understand the limits of complex functions.