Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the limits. $$ \begin{array}{l} \underline{\phantom{xxx}} \lim _{x \rightarrow \infty}\left(\sqrt{2 x^{2}+3}-\sqrt{2 x^{2}-5}\right) . \text { Hint: } \quad \text { Multiply and }\\\ \text { divide by } \sqrt{2 x^{2}+3}+\sqrt{2 x^{2}-5} \end{array} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Limit Expression

We need to find the limit as \( x \) approaches infinity for the expression \( \lim_{x \rightarrow \infty}\left(\sqrt{2 x^{2}+3}-\sqrt{2 x^{2}-5}\right) \). It's a difference of square roots, which suggests the rationalization method could be useful.

Step 2: Multiply and Divide by the Conjugate

To simplify the expression, multiply and divide by the conjugate: \( \sqrt{2 x^{2}+3}+\sqrt{2 x^{2}-5} \). This is a common technique to handle differences of square roots.

Step 3: Simplify the Expression

Compute the product: \[\left(\sqrt{2 x^{2}+3}-\sqrt{2 x^{2}-5}\right) \cdot \frac{\sqrt{2 x^{2}+3}+\sqrt{2 x^{2}-5}}{\sqrt{2 x^{2}+3}+\sqrt{2 x^{2}-5}} = \frac{(2x^2 + 3) - (2x^2 - 5)}{\sqrt{2 x^{2}+3}+\sqrt{2 x^{2}-5}}.\]Simplifying the numerator gives \(8\), leading to: \( \frac{8}{\sqrt{2 x^{2}+3}+\sqrt{2 x^{2}-5}} \).

Step 4: Evaluate the Limit of the Simplified Expression

As \( x \to \infty \), evaluate the limit of \( \frac{8}{\sqrt{2 x^{2}+3}+\sqrt{2 x^{2}-5}} \). The denominators \( \sqrt{2 x^{2}+3} \) and \( \sqrt{2 x^{2}-5} \) approach \( \sqrt{2}x \) as \( x \to \infty \). Thus, the expression becomes \( \sqrt{2}x + \sqrt{2}x = 2\sqrt{2}x \).The limit is \( \lim_{x \to \infty} \frac{8}{2\sqrt{2}x} = 0 \).

Key Concepts

Infinity in Calculus

In calculus, the concept of infinity is often used when evaluating limits. It helps us understand the behavior of functions as they grow larger and larger or decrease indefinitely. When we say a variable like \( x \) approaches infinity, symbolized as \( x \to \infty \), we are essentially interested in what the function does as \( x \) becomes very large.

This idea of approaching infinity allows us to simplify and approximate complicated expressions. It gives insight into trends rather than specific values, which is crucial when dealing with limits. In our exercise, \( x \) approaches infinity within a square root function, leading us to consider how square roots behave with infinitely large inputs. Understanding how terms like \( \sqrt{2x^2} \) can dominate and simplify other terms as \( x \) becomes large is key to mastering limits involving infinity.

Rationalizing Expressions

Rationalizing expressions is a powerful algebraic technique often used when dealing with limits involving square roots. It involves multiplying the expression by a conjugate to reduce radicals or complex terms. Conjugates are pairs like \( \sqrt{2x^2+3} + \sqrt{2x^2-5} \), which can help simplify expressions by eliminating square roots during multiplication.

In our case, to handle the difference of square roots , we multiply by this conjugate. This technique transforms the original complex expression into a simpler form through subtraction: \( (2x^2 + 3) - (2x^2 - 5) \), simplifying the limit process. Rationalizing is particularly useful because it allows us to manage indeterminate forms and make calculations feasible.

Square Roots in Calculus

Square roots can make calculus problems complex, particularly in limits. When square roots involve variables that approach infinity, recognizing dominant terms becomes essential. For instance, in \( \sqrt{2x^2+3} \) and \( \sqrt{2x^2-5} \), as \( x \) grows, the \( 2x^2 \) term largely dictates the behavior of the function.

By focusing on these leading terms, we simplify evaluation: the smaller terms (\(+3\) and \(-5\)) become negligible. The square root expressions simplify to approximate forms like \( \sqrt{2}x \). This insight helps in calculating limits, as it reveals how vast numbers influence function values, allowing you to skillfully navigate through the apparent complexity of square roots in limits.