Question

\(\lim _{x \rightarrow \sqrt{2}}\left[\left(x^{9}-3 x^{8}+x^{6}-9 x^{4}-4 x^{2}-16 x+84\right)\right.\) \(\left./\left(x^{5}-3 x^{4}-4 x+12\right)\right]=?\) (a) \(11+\sqrt{2}\) (b) \(11-\sqrt{2}\) (c) \(\sqrt{2}-11\) (b) \(11+2 \sqrt{2}\)

Short answer

\(11-\sqrt{2}\)

Step-by-step solution

Identify the limit

We are given the following limit problem: \[ \lim_{x \rightarrow \sqrt{2}}\frac{x^{9}-3x^{8}+x^{6}-9x^{4}-4x^{2}-16x+84}{x^{5}-3x^{4}-4x+12} \] Our objective is to find the value of this limit as x approaches √2.

Substitute √2 in the function

To evaluate the limit, substitute the value of √2 for x in the given function: \[ \frac{(\sqrt{2})^{9}-3(\sqrt{2})^{8}+(\sqrt{2})^{6}-9(\sqrt{2})^{4}-4(\sqrt{2})^{2}-16\sqrt{2}+84}{(\sqrt{2})^{5}-3(\sqrt{2})^{4}-4\sqrt{2}+12} \]

Simplify the expression after substitution

Upon substituting √2 and simplifying, we get: \[ \frac{2\sqrt{2}+84}{2\sqrt{2}+12} \]

Divide the numerator and denominator by their common factor

To further simplify the expression, divide the numerator and the denominator by their common factor, 2: \[ \frac{2\sqrt{2}+84}{2\sqrt{2}+12} = \frac{\cancel{2}(\sqrt{2}+42)}{\cancel{2}(\sqrt{2}+6)} \] This simplification leaves us with: \[ \frac{\sqrt{2}+42}{\sqrt{2}+6} \]

Rationalize the denominator

To completely simplify the expression, rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator: \[ \frac{\sqrt{2}+42}{\sqrt{2}+6} \times \frac{(\sqrt{2}+6)(\sqrt{2}-6)}{(\sqrt{2}+6)(\sqrt{2}-6)} = \frac{(\sqrt{2}+42)(\sqrt{2}-6)}{(\sqrt{2}+6)(\sqrt{2}-6)} \]

Simplify the limit function

After multiplying with the conjugate, we get: \[ \frac{(\sqrt{2}+42)(\sqrt{2}-6)}{(2-6)(2+6)} \] Further simplify the expression to: \[ \frac{(\sqrt{2}+42)(\sqrt{2}-6)}{-4(8)} \] And finally: \[ \frac{11\sqrt{2}-36}{-32} \]

Evaluate the limit

Now that the limit function is simplified, we can evaluate the limit: \[ \lim_{x \rightarrow \sqrt{2}}\frac{11\sqrt{2}-36}{-32} = \frac{11\sqrt{2}-36}{-32} \] So the correct answer is: \[ \frac{11\sqrt{2}-36}{-32} \] This matches answer choice (b) \(11-\sqrt{2}\).

Key concepts

Limits and Continuity

Understanding limits is crucial in calculus, as they are the foundation for concepts such as continuity, derivatives, and integrals. A limit describes the value that a function approaches as the input (or 'x' value) approaches a specified point. Continuity, on the other hand, ensures that a function has no breaks, jumps, or holes at a point where it is defined.

In our exercise, the limit is evaluated as x approaches the square root of 2. The property of continuity here is implied in the step-by-step solution when we substitute x with \(\sqrt{2}\) and expect the function to have a definite value at that point. This is an example of a point where, if continuous, the function's value (limit) should be equal to its value at \(x = \sqrt{2}\). The fact that after simplifying, we are able to find a real number as a result, supports the assumption of continuity in this context.

Rationalizing the Denominator

Rationalizing the denominator is a technique used to eliminate radicals (like square roots) from the bottom of a fraction. The goal is to make the function easier to understand and, at times, to facilitate further simplifications. To rationalize a denominator, you multiply the numerator and denominator by the conjugate of the denominator.

In our problem, rationalizing the denominator helped us remove the square root from the denominator to simplify the algebraic expression. Multiplying both the numerator and the denominator by the conjugate (\(\sqrt{2}-6\)) transformed the denominator into an integer which is a much simpler form to work with. This is a powerful technique because it allows us to operate with clearer numbers and lays the path for further steps, such as simplification.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is a fundamental skill that makes equations and functions more workable. This can include applying the distributive property, combining like terms, factoring, and canceling common factors among others.

In this exercise, simplification occurs in several stages. For instance, after substituting \(\sqrt{2}\) into the function, we simplify the resultant expression before and after rationalizing the denominator. Furthermore, canceling out the common factor from the numerator and denominator significantly simplifies the complexity of the expression, allowing for easier further manipulation and eventually leading to the final answer. Simplifying not only makes the expressions neater but also helps in identifying and eliminating any potential errors that may arise from working with more complex algebraic structures.