Question

If \(\operatorname{Lt}_{x \rightarrow m} \frac{x^{3}-m^{3}}{x-m}=3\), then find the number of possible values of \(m\). (1) 0 (2) 2 (3) 1 (4) 3

Short answer

Answer: 1

Step-by-step solution

Rewrite the limit expression using the difference of cubes formula

We are given the limit expression: \(\operatorname{Lt}_{x \rightarrow m} \frac{x^{3}-m^{3}}{x-m}\). First, rewrite the numerator using the difference of cubes formula: \(x^3 - m^3 = (x - m)(x^2 + xm + m^2)\).

Simplify the limit expression

Now, substitute this expression back into the original limit expression and simplify it: \(\operatorname{Lt}_{x \rightarrow m} \frac{(x - m)(x^2 + xm + m^2)}{x-m}\) Cancel out the \((x - m)\) terms in the numerator and denominator: \(\operatorname{Lt}_{x \rightarrow m} (x^2 + xm + m^2)\)

Evaluate the limit

Now, evaluate the limit by substituting \(x\) with \(m\): \((m^2 + m\cdot m + m^2) = (2m^2 + m^3)\)

Set the limit equal to the given value and solve for m

We are given that the limit equals 3, so: \(2m^2 + m^3 = 3\) Rearrange to have all terms on one side: \(m^3 + 2m^2 - 3 = 0\) Now, analyze the values of the polynomial around different integers to check for the roots. For example, since -1 is a root, m could be 0, 1, or -2.

Determine the number of possible values

Experiment with these values to see if any of them satisfy the given equation: If m = 0: \((0)^3 + 2(0)^2 - 3 = -3 ≠ 0\) If m = 1: \((1)^3 + 2(1)^2 - 3 = 0\) If m = -2: \((-2)^3 + 2(-2)^2 - 3 = -8 + 8 - 3 = -3 ≠ 0\) Only m = 1 satisfies the given equation, so there is only 1 possible value of m. Thus, the correct answer is (3) 1.

Key concepts

Difference of Cubes

The difference of cubes is an important algebraic identity that can help simplify complex expressions. It allows us to express the difference between two cubes as a product. This comes in particularly handy when evaluating limits involving cubic expressions. The formula for the difference of cubes is:\[x^3 - m^3 = (x - m)(x^2 + xm + m^2) \]Here, the expression on the left represents the difference between two cubes, and the expression on the right is the factored form. By factoring, we can cancel terms in equations like the limit given in the example problem.

Using this identity is especially useful because it breaks down a more complex cubic polynomial into simpler linear and quadratic factors. This can often make solving equations, like polynomial equations in limit problems, easier and more intuitive.

Polynomial Equations

Polynomial equations involve expressions that are equated to zero with one or more terms. These terms are composed of variables raised to various powers. They play a key role in calculus and algebra. A cubic polynomial, such as the one from our exercise—\(m^3 + 2m^2 - 3 = 0\)—requires roots finding techniques to solve.

In mathematically evaluating polynomial equations, you might use methods such as:

• Factoring: Breaking down complex expressions into simpler factors.
• Substitution: Testing potential roots by substituting values directly into the polynomial to check if it equals zero.
• Graphing: Visualizing the polynomial to find where it crosses the axis.

In the exercise, we resolved the equation through trial and error by testing values like \( m = 0, 1, -2 \) to determine which satisfies the polynomial. This trial involves substituting each possible root back into the equation to see if the left-hand side equals zero.

Limit Evaluation Techniques

When evaluating limits, especially those involving polynomials, special techniques can simplify the process. The given exercise required evaluating \( \operatorname{Lt}_{x \rightarrow m} \frac{x^3 - m^3}{x - m} \). After simplifying with the difference of cubes, we used the direct substitution technique once the expression was simplified.

Here are some common techniques for evaluating limits:

• Direct Substitution: If directly substituting the approaching value doesn't result in an indeterminate form, this is usually the go-to method.
• Simplification or Factoring: Simplifying the expression, like using identities to cancel terms, often allows direct substitution afterward.
• Limit Laws: Using rules such as the sum, product, and quotient laws to divide and conquer complex parts.

In the example, simplifying by canceling \( (x - m) \) terms enabled straightforward evaluation by plugging in \( x = m \). This yielded the polynomial equation which we then solved for possible values of \( m \).