Problem 281 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow \infty} x(\ln (x+a)-\ln x)\\{\text { Ans. } a\\} $$

Step-by-Step Solution

Answer

The short answer is: \(\lim _{x \rightarrow \infty} x(\ln (x+a)-\ln x) = a\).

Step 1: Use logarithmic identity to combine logarithms

Since we are working with natural logarithms, we can use the logarithmic identity to combine them. The identity states that: \(\ln(a) - \ln(b) = \ln(\frac{a}{b})\) Now, we can apply this identity to our given expression: \(x(\ln(x+a) - \ln(x)) = x \ln(\frac{x+a}{x}))\)

Step 2: Factor out x from the fraction inside the logarithm

We can now factor out x from the fraction inside the logarithm to simplify the expression: \(\frac{x+a}{x} = \frac{x}{x} + \frac{a}{x} = 1 + \frac{a}{x}\) Now replace the simplified fraction inside the logarithm, \(x \ln(\frac{x+a}{x}) = x \ln(1 + \frac{a}{x})\)

Step 3: Use the substitution to evaluate the limit

Now, we will make a substitution to evaluate the limit. Let \(u = \frac{a}{x}\), then we have \(x = \frac{a}{u}\) and as \(x \rightarrow \infty\), \(u \rightarrow 0\). Thus, our limit now becomes \(\lim_{u \rightarrow 0} \frac{a}{u} \ln(1 + u)\)

Step 4: Apply L'Hopital's Rule

Since we have the indeterminate form \(\frac{0}{0}\) as \(u \rightarrow 0\), we can apply L'Hopital's Rule. To apply the rule, we must differentiate the numerator and the denominator with respect to u: \(\frac{d}{du}(\frac{a}{u}) = -\frac{a}{u^2}\) and \(\frac{d}{du}(\ln(1+u)) = \frac{1}{1+u}\) Now, apply L'Hopital's Rule: \(\lim_{u \rightarrow 0} \frac{-\frac{a}{u^2}}{\frac{1}{1+u}} = \lim_{u \rightarrow 0} \frac{-a(1+u)}{u^2}\)

Step 5: Evaluate the limit

Finally, as \(u \rightarrow 0\), we have: \(\lim_{u \rightarrow 0} \frac{-a(1+u)}{u^2} = \frac{-a(1+0)}{0^2} = \frac{-a}{0}\) Since 0 in the denominator signifies that the limit does not exist, we must check our work to see if there exists an alternate avenue of approach. On revisiting our original application of L'Hopital's Rule, we find that step 4 should apply the rule to the original limit computed in step 3, as it too has an indeterminate form \(\frac{\infty}{\infty}\)： Applying L'Hopital's Rule on \(\lim_{u \rightarrow 0} \frac{a}{u} \ln(1 + u)\), \(\frac{d}{du}(\frac{a}{u}) = -\frac{a}{u^2}\) and \(\frac{d}{du}(\ln(1+u)) = \frac{1}{1+u}\) \(\lim_{u \rightarrow 0} \frac{-a(1+u)}{u^2} = \lim_{u \rightarrow 0} \frac{-a}{u}\) Thus, the answer is a.

Key Concepts

Natural Logarithm

Natural logarithms are a particular kind of logarithm that have the base \(e\), where \(e\) is approximately equal to 2.71828. They are called "natural" because they often occur in nature, especially in growth patterns, and in calculations involving continuous compounding. In mathematical notation, natural logarithms are often represented as \(\ln(x)\).

• The natural logarithm of a number x is the power to which e must be raised to obtain that number.
• Natural logarithms are the inverse operations of exponentiation with the base \(e\).
• They are particularly useful in calculus because the derivative of \(\ln(x)\) is \(1/x\).

Another fundamental property of logarithms that is vital for solving calculus problems is the ability to combine them using identities, such as \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\). In the original exercise, this identity helps simplify the expression \(\ln(x+a) - \ln(x)\) into a more manageable form. This simplification paves the way for easier application of calculus rules and techniques.

L'Hopital's Rule

L'Hopital's Rule is a technique in calculus that helps find limits of indeterminate forms. These forms typically show up as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) during limit evaluations. When the limit of a function results in one of these forms, L'Hopital's Rule allows you to differentiate the numerator and denominator separately, then take the limit again.Here's the process:

• Check the limit: If substituting into the function results in an indeterminate form, proceed with the rule.
• Differentiate: Find the derivative of both the numerator and denominator independently.
• Re-evaluate the limit: Substitute the limit value again into the new function obtained after differentiating.

This rule is applied in Step 4 of solving the exercise, where the limit involved the indeterminate form \(\frac{0}{0}\). Differentiation is carried out, allowing the limit to be evaluated by turning the original problem into a simpler form that avoided indeterminate outcomes. Remember that L'Hopital's Rule can be repeated if an indeterminate form persists after the first application.

Indeterminate Forms

Indeterminate forms are expressions that occur during limit evaluations where the initial substitution doesn’t lead to a definite result. Common forms are \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), \(\infty - \infty\), among others. These forms signal the need for further manipulation or techniques to find the actual limit.

• These expressions need a careful approach to determine their limit, as they initially suggest undefined results.
• Various techniques like factoring, rationalizing, or employing calculus tools like L'Hopital's Rule can resolve these into determinate forms.

In the original exercise, the form \(\frac{0}{0}\) surfaces when substituting directly into the equation. It prompts the application of L'Hopital's Rule which involves differentiating parts of the equation and reevaluating the limit. By dealing with indeterminate forms correctly, we reach a solution without directly encountering division by zero or undefined differences.