Understanding the Limit Problem

We are asked to find the limit of the function \( \dfrac{x+x^2+x^3-3}{x-1} \) as \( x \) approaches 1. This is a common type of limit problem encountered in calculus.

Let's first substitute \( x=1 \) into the expression to see what we get:

\[ \dfrac{1+1^2+1^3-3}{1-1} = \dfrac{1+1+1-3}{0} = \dfrac{3-3}{0} = \dfrac{0}{0} \]

Since we get the indeterminate form \( \dfrac{0}{0} \), we need to use a different method to evaluate the limit. Common methods for indeterminate forms include factorization or L'Hôpital's Rule.

Solving the Limit using L'Hôpital's Rule

L'Hôpital's Rule can be applied when evaluating limits that result in indeterminate forms like \( \dfrac{0}{0} \) or \( \dfrac{\infty}{\infty} \). The rule states that if \( \lim_{x \to c} \dfrac{f(x)}{g(x)} \) is of the form \( \dfrac{0}{0} \) or \( \dfrac{\infty}{\infty} \), then \( \lim_{x \to c} \dfrac{f(x)}{g(x)} = \lim_{x \to c} \dfrac{f'(x)}{g'(x)} \), provided the latter limit exists.

In our case, let \( f(x) = x+x^2+x^3-3 \) and \( g(x) = x-1 \). We need to find the derivatives of \( f(x) \) and \( g(x) \) with respect to \( x \).

• Derivative of \( f(x) \): \( f'(x) = \dfrac{d}{dx}(x+x^2+x^3-3) = \dfrac{d}{dx}(x) + \dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(x^3) - \dfrac{d}{dx}(3) \)
• \( f'(x) = 1 + 2x + 3x^2 - 0 = 1+2x+3x^2 \)
• Derivative of \( g(x) \): \( g'(x) = \dfrac{d}{dx}(x-1) = \dfrac{d}{dx}(x) - \dfrac{d}{dx}(1) \)
• \( g'(x) = 1 - 0 = 1 \)

Now, applying L'Hôpital's Rule:

\[ \displaystyle\lim_{x \rightarrow 1} \dfrac{x+x^2+x^3-3}{x-1} = \lim_{x \rightarrow 1} \dfrac{1+2x+3x^2}{1} \]

Now, substitute \( x=1 \) into the new expression:

\[ \dfrac{1+2(1)+3(1)^2}{1} = \dfrac{1+2+3(1)}{1} = \dfrac{1+2+3}{1} = \dfrac{6}{1} = 6 \]

Solving the Limit using Factorization

Alternatively, since substituting \( x=1 \) into the numerator \( x+x^2+x^3-3 \) gives 0, this means that \( (x-1) \) is a factor of the numerator polynomial. We can rewrite the numerator to factor out \( (x-1) \).

Numerator: \( x+x^2+x^3-3 \)

We can rewrite the terms by subtracting and adding 1 strategically:

\[ (x-1) + (x^2-1) + (x^3-1) \]

Now, we can factor each part using the difference of squares formula \( a^2-b^2 = (a-b)(a+b) \) and the difference of cubes formula \( a^3-b^3 = (a-b)(a^2+ab+b^2) \):

• \( x-1 = (x-1) \)
• \( x^2-1 = (x-1)(x+1) \)
• \( x^3-1 = (x-1)(x^2+x+1) \)

Substitute these factored forms back into the numerator expression:

\[ (x-1) + (x-1)(x+1) + (x-1)(x^2+x+1) \]

Now, factor out the common term \( (x-1) \):

\[ (x-1) [1 + (x+1) + (x^2+x+1)] \]

Simplify the expression inside the square brackets:

\[ (x-1) [1 + x+1 + x^2+x+1] = (x-1) [x^2 + (x+x) + (1+1+1)] = (x-1) [x^2+2x+3] \]

So the numerator \( x+x^2+x^3-3 \) is equal to \( (x-1)(x^2+2x+3) \).

Now, substitute this back into the limit expression:

\[ \displaystyle\lim_{x \rightarrow 1} \dfrac{(x-1)(x^2+2x+3)}{x-1} \]

For \( x \neq 1 \), the term \( (x-1) \) is non-zero, so we can cancel it from the numerator and the denominator:

\[ \displaystyle\lim_{x \rightarrow 1} (x^2+2x+3) \]

Now, we can substitute \( x=1 \) into this simplified expression:

\[ 1^2 + 2(1) + 3 = 1 + 2 + 3 = 6 \]

Both methods yield the same result, which is 6.

Final Answer Determination

Based on the calculations using both L'Hôpital's Rule and factorization, the limit \( \displaystyle\lim_{x \rightarrow 1} \dfrac{x+x^2+x^3-3}{x-1} \) is equal to 6.

Revision Table - Limit Evaluation

Additional Information - Related Calculus Concepts

Understanding limits is fundamental in calculus. Limits are used to define continuity, derivatives, and integrals.

• Continuity: A function \( f(x) \) is continuous at a point \( c \) if \( \lim_{x \to c} f(x) = f(c) \). In our problem, the original function is not continuous at \( x=1 \) because the denominator is zero. However, the limit still exists.
• Derivatives: The derivative of a function at a point is defined using a limit. \( f'(c) = \lim_{h \to 0} \dfrac{f(c+h)-f(c)}{h} \). L'Hôpital's Rule directly uses the concept of derivatives to evaluate limits of quotients.
• Polynomial Functions: Polynomials are continuous everywhere. The techniques used (factorization, L'Hôpital's Rule involving polynomial derivatives) are standard for problems involving rational functions (ratios of polynomials).