Question

Calculate the following limits using the factorization formula \(x^{n}-a^{n}=(x-a)\left(x^{n-1}+a x^{n-2}+a^{2} x^{n-3}+\cdots+a^{n-2} x+a^{n-1}\right)\) where \(n\) is a positive integer and a is a real number. $$\lim _{x \rightarrow 2} \frac{x^{5}-32}{x-2}$$

Short answer

Question: Find the limit of the given expression as x approaches 2: $$\lim _{x \rightarrow 2} \frac{x^{5}-32}{x-2}$$ Answer: 80

Step-by-step solution

Apply the factorization formula

Since we have the expression in the form of \(x^n - a^n\) and the factorization formula provided, we can directly apply the formula to simplify the given expression: Formula: \(x^{n}-a^{n}=(x-a)\left(x^{n-1}+a x^{n-2}+a^{2} x^{n-3}+\cdots+a^{n-2} x+a^{n-1}\right)\) where \(n=5\) and \(a=2\): $$x^{5}-32=(x-2)\left(x^{4}+2x^{3}+2^2x^{2}+2^3x+2^4\right)$$

Substitute the factorization in the given limit

Now that we have the factorization, substitute it into the given limit expression: $$\lim _{x \rightarrow 2} \frac{(x-2)\left(x^{4}+2x^{3}+4x^{2}+8x+16\right)}{x-2}$$

Simplify the expression

Since we have a common factor of \((x-2)\) in both the numerator and denominator, we can cancel them: $$\lim _{x \rightarrow 2} \left(x^{4}+2x^{3}+4x^{2}+8x+16\right)$$

Evaluate the limit

Now, we can simply substitute the value \(x=2\) into the simplified expression to evaluate the limit: $$\lim _{x \rightarrow 2} \left(x^{4}+2x^{3}+4x^{2}+8x+16\right) = (2)^{4}+2(2)^{3}+4(2)^{2}+8(2)+16$$ $$=16+16+16+16+16=80$$ Thus, the limit of the given expression as x approaches 2 is 80: $$\lim _{x \rightarrow 2} \frac{x^{5}-32}{x-2}=80$$

Key concepts

Factorization Formula

The factorization formula is a powerful tool in algebra that helps in breaking down complex polynomials into more manageable parts. It is particularly useful when encountering limits in calculus where direct substitution would result in an undefined form. A common type of factorization formula used in calculus is the difference of nth powers, given as \(x^{n} - a^{n} = (x - a)(x^{n-1} + a x^{n-2} + a^{2} x^{n-3} + \cdots + a^{n-2} x + a^{n-1})\). This formula applies when you're trying to find the limit of a function involving a difference of two terms raised to the same power, as in the exercise.
When used appropriately, the formula can simplify a polynomial expression by removing a common factor, and this simplification can often lead to easier limit evaluation. The resulting terms after factorization can then be used to evaluate the limit without the indeterminate form that might initially arise. In our example, the common factor \((x-2)\) was removed, paving the way for a straightforward substitution to find the limit value.

Evaluating Limits

Evaluating limits is a fundamental concept in calculus, involving finding the value that a function approaches as the input approaches a certain point. The challenge often arises when substitution of the point into the function leads to an undefined expression, such as 0/0. To navigate this, factorization, as shown in the previous section, can be used to simplify the expression.
Once an expression is simplified, evaluating the limit often becomes as straightforward as direct substitution. If the simplified function is continuous at the point of interest, the limit is simply the value of the function at that point. As illustrated in the provided exercise, after applying the factorization formula and canceling the common factor, the limit was successfully evaluated by plugging in the value of \(x\).

Polynomial Functions

Polynomial functions are algebraic expressions consisting of variables and coefficients, structured in terms of powers of the variable with non-negative integer exponents. They are widely recognized for their simplicity and smooth behavior. Examples include quadratic functions (\(ax^2+bx+c\)), cubic functions (\(ax^3+bx^2+cx+d\)), and higher power polynomials.
A distinctive feature of polynomials is their continuity over the entire set of real numbers. This means that we can evaluate their limits by direct substitution. In the exercise, after simplifying the polynomial by factorization and cancellation, we were left with a polynomial expression that we could evaluate using substitution, owing to this specific feature of polynomial functions.

Limits of Functions

Limits capture the behavior of a function as the input variable approaches a particular value. Understanding limits is crucial in calculus because they underpin the concepts of continuity, derivatives, and integrals. For functions that are not continuous at a point, direct substitution is not possible, and alternative strategies like factorization must be employed.
In the provided example, the limit evaluation was initially problematic due to a division by zero situation. However, after applying the factorization formula and simplifying the expression, we were able to eliminate this problem and easily find the limit of the function as \(x\) approached 2. The end result, the limit of the function, tells us about the behavior of the function near that specific point, without the need to evaluate the function exactly at that point.