Problem 14 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

\(\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}+b^{2}}}{\sqrt{x^{2}+c^{2}}-\sqrt{x^{2}+d^{2}}}=\) (a) \(\frac{a^{2}-b^{2}}{c^{2}-d^{2}}\) (b) \(\frac{a^{2}+b^{2}}{c^{2}-d^{2}}\) (c) \(\frac{a^{2}+b^{2}}{c^{2}+d^{2}}\) (d) None of these

Step-by-Step Solution

Answer

The limit evaluates to option (a): \( \frac{a^{2}-b^{2}}{c^{2}-d^{2}} \).

Step 1: Simplify the Expression

Given the limit \( \lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}+b^{2}}}{\sqrt{x^{2}+c^{2}}-\sqrt{x^{2}+d^{2}}} \), we start by rationalizing both the numerator and denominator. This is done by multiplying and dividing by their respective conjugates.

Step 2: Rationalize the Numerator

Multiply and divide the numerator \( \sqrt{x^{2}+a^{2}}-\sqrt{x^{2}+b^{2}} \) by its conjugate \( \sqrt{x^{2}+a^{2}}+\sqrt{x^{2}+b^{2}} \). This gives us: \[\frac{(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}+b^{2}})(\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}+b^{2}})}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}+b^{2}}} = \frac{(x^2+a^2)-(x^2+b^2)}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}+b^{2}}} = \frac{a^2-b^2}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}+b^{2}}} \]

Step 3: Rationalize the Denominator

Similarly, multiply and divide the denominator \( \sqrt{x^{2}+c^{2}}-\sqrt{x^{2}+d^{2}} \) by its conjugate \( \sqrt{x^{2}+c^{2}}+\sqrt{x^{2}+d^{2}} \). This gives us: \[\frac{(\sqrt{x^{2}+c^{2}}-\sqrt{x^{2}+d^{2}})(\sqrt{x^{2}+c^{2}}+\sqrt{x^{2}+d^{2}})}{\sqrt{x^{2}+c^{2}}+\sqrt{x^{2}+d^{2}}} = \frac{(x^2+c^2)-(x^2+d^2)}{\sqrt{x^{2}+c^{2}}+\sqrt{x^{2}+d^{2}}} = \frac{c^2-d^2}{\sqrt{x^{2}+c^{2}}+\sqrt{x^{2}+d^{2}}} \]

Step 4: Evaluate the Limit

Substitute the simplified forms from Steps 2 and 3 back into the original expression:\[ \lim _{x \rightarrow \infty} \frac{\frac{a^2-b^2}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}+b^{2}}}}{\frac{c^2-d^2}{\sqrt{x^{2}+c^{2}}+\sqrt{x^{2}+d^{2}}}} = \lim _{x \rightarrow \infty} \frac{(a^2-b^2)(\sqrt{x^{2}+c^{2}}+\sqrt{x^{2}+d^{2}})}{(c^2-d^2)(\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}+b^{2}})} \] As \( x \to \infty \), the dominant terms in the square roots are \( x \), so all square root terms approach \( x \). Therefore, the limit simplifies to:\[ \frac{(a^2-b^2)x}{(c^2-d^2)x} = \frac{a^2-b^2}{c^2-d^2} \]

Step 5: Select the Correct Answer

The evaluation of the limit produces \( \frac{a^2-b^2}{c^2-d^2} \). Hence, the correct choice aligns with option (a).

Key Concepts

Rationalization

In calculus, rationalization is a critical technique used to simplify expressions, particularly those involving square roots. At its core, rationalization involves multiplying a fraction by a suitable term that eliminates irrational components from its denominator. In the given problem, we have an expression involving square roots: \[ \frac{\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}+b^{2}}}{\sqrt{x^{2}+c^{2}}-\sqrt{x^{2}+d^{2}}} \]Both the numerator and the denominator contain square roots. To make this expression more manageable, we use their conjugates. The conjugate of a binomial with square root terms is simply the same terms but with an opposite sign between them. For example, if you have \( \sqrt{x^2+a^2} - \sqrt{x^2+b^2} \), its conjugate is \( \sqrt{x^2+a^2} + \sqrt{x^2+b^2} \). Multiplying by a conjugate utilizes the difference of squares formula:

• \( (a-b)(a+b) = a^2 - b^2 \)

Applying this to both the numerator and the denominator converts the irrational components in the fraction into more straightforward polynomials, as shown in steps 2 and 3 of the solution. Rationalization simplifies the evaluation of limits, paving the way for easier limit calculations.

Infinite Limits

Infinite limits involve evaluating how a function behaves as its variable approaches infinity \( (x \to \infty) \) or negative infinity \( (x \to -\infty) \). When dealing with infinite limits, especially those involving radicals, it is crucial to understand which terms within expressions dominate as the variable grows large. In this exercise, the key is to comprehend how:\[ \lim_{x \to \infty} \frac{\sqrt{x^2+a^2} - \sqrt{x^2+b^2}}{\sqrt{x^2+c^2} - \sqrt{x^2+d^2}} \]behaves. Understanding infinite limits helps simplify the expression by focusing on dominant terms, resulting in finite limits as opposed to an undefined infinite value. After rationalization, we see that as \( x \to \infty \), the dominant behavior in each square root term is driven largely by \( x^2 \). This means contributions from \( a^2, b^2, c^2, \) and \( d^2 \) become negligible. By focusing on these limits, we see how \[ \lim_{x \to \infty} \frac{a^2-b^2}{c^2-d^2} \]is reached, which is a finite value. This transition highlights the power of limit evaluation in simplifying seemingly complex expressions as they approach infinity.

Dominant Terms

When computing limits, especially as \( x \to \infty \), it's vital to identify dominant terms in algebraic expressions. Dominant terms are those that grow significantly faster than other terms as the variable increases in magnitude. In the expression:\[ \sqrt{x^2+a^2} \] As \( x \to \infty \), the term \( x^2 \) inside the square root becomes the dominant component. This is because \( \sqrt{x^2+a^2} \approx \sqrt{x^2} = x\), making \( a^2 \) insignificant in comparison.When evaluating limits, dominant terms help strip down expressions to their essential components. This is achieved by:

• Recognizing the largest power of \( x \) in polynomial expressions or under square roots.
• Approximating other terms as negligible if they do not match the growth rate of the dominant terms.

By focusing on dominant terms, we simplify the problem to a manageable level, allowing us to resolve expressions like our initial limit down to:\[ \frac{a^2-b^2}{c^2-d^2} \]as \( x \to \infty \). This reduction hinges on the dominant characteristic of linear terms like \( x \) in radicals as they tend to simplify complex expressions to their core behavior under limits.