Question

\(\lim _{\mathrm{x} \rightarrow 1}\left[\left\\{\left[{ }^{3} \sum_{\mathrm{i}=1}(\mathrm{x}+\mathrm{i})^{2}\right]-29\right\\} /(\mathrm{x}-1)\right]=?\) (a) 9 (b) 12 (c) 18 (d) 30

Short answer

The limit of the given expression as x approaches 1 is \(18\). Thus, the answer is (c) 18.

Step-by-step solution

Evaluate the sum

First, we will evaluate the summation. In the given expression, we have a summation with an upper limit of 3: \[\sum_{i=1}^3 (x+i)^2\] To evaluate this sum, we can plug in i=1, i=2, and i=3 into the expression and add the results: \[(x+1)^2 + (x+2)^2 + (x+3)^2\]

Simplify the sum

Now let's simplify the sum by expanding the square terms: \[(x^2 + 2x + 1) + (x^2 + 4x + 4) + (x^2 + 6x + 9)\] Combine like terms to further simplify the sum: \[3x^2 + 12x + 14\]

Substitute the sum back into the expression

Now that we have simplified the sum, we can substitute it back into the given expression: \[\lim_{x\to1}\frac{[(3x^2 + 12x + 14) - 29]}{(x -1)}\]

Simplify the expression

Next, we will simplify the expression within the limit: \[\lim_{x\to1}\frac{(3x^2 + 12x - 15)}{(x -1)}\]

Factor the expression

To evaluate this limit, we need to factor the expression in the numerator: \[\lim_{x\to1}\frac{3(x^2 + 4x - 5)}{(x -1)}\] Now factor the quadratic expression in the numerator: \[\lim_{x\to1}\frac{3(x-1)(x+5)}{(x -1)}\]

Simplify and evaluate the limit

Now we can apply the cancellation of the common factor (x - 1) in the numerator and denominator: \[\lim_{x\to1} 3(x+5)\] Now substitute x = 1 into the simplified expression and evaluate the limit: \[3(1+5) = 3 \cdot 6 = 18\] Since the limit as x approaches 1 is equal to 18, the correct answer is (c) 18.

Key concepts

Summation

Summation involves adding up a sequence of numbers, which can be represented by the sum symbol \( \sum \). In our exercise, we have a limit that involves a summation notation \( \sum_{i=1}^3 (x+i)^2 \). This tells us to substitute different values of \( i \) starting from 1 to 3 into the expression \( (x+i)^2 \) and then add all those values together.

Here's how it works:

• First, with \( i = 1 \), compute \( (x+1)^2 \).
• Next, with \( i = 2 \), compute \( (x+2)^2 \).
• Finally, with \( i = 3 \), compute \( (x+3)^2 \).

These results are summed together, which is a practical application of the summation concept. It's a fundamental part of calculus that allows us to handle series and sequences efficiently.

Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial expression into simpler "factors" that, when multiplied, give the original expression. In our solution, factoring is crucial, as it helps simplify the limits question.

In our exercise, after simplifying the summation, we get to the expression \( 3x^2 + 12x - 15 \). To factor this, observe that we can pull out any common factor first. Here, we take out a 3:

\[ 3(x^2 + 4x - 5) \]
We then need to factor the quadratic expression \( x^2 + 4x - 5 \) further. We look for two numbers that multiply to \(-5\) and add to \(4\). The suitable pair is \(+5\) and \(-1\), so we can rewrite it as:\[ (x+5)(x-1) \].

Factoring is a powerful tool in algebra and calculus for simplifying expressions and solving equations.

Evaluating Limits

Evaluating limits allows us to find the value that a function approaches as the input approaches a certain point. In problems involving limits and polynomials, like our current problem, simplification through factoring is often key to evaluate that limit effectively.

After transforming our problem through summation and factoring, we end up with:
\[ \lim_{x\to1}\frac{3(x-1)(x+5)}{(x-1)} \]
When \( x \to 1 \), the term \(x - 1\) in the numerator and denominator can be canceled, simplifying the limit to:
\[ \lim_{x\to1} 3(x+5) \]
We then directly substitute \( x = 1 \), yielding a result of \( 18 \). This demonstrates how evaluating limits involves careful algebraic manipulation to reach a point where direct substitution is possible.

Polynomial Expressions

Polynomial expressions like \( 3x^2 + 12x - 15 \) are a cornerstone of calculus and algebra. These expressions consist of terms where a variable is raised to a non-negative integer power and scaled by coefficients.

In our exercise, understanding the structure of polynomials was crucial in each step, from simplifying the summation to factoring it correctly. Polynomials are typically managed through:

• Manipulating the terms using algebraic operations.
• Factoring them to simplify expressions or solve equations.
• Substituting values to evaluate limits or solve for specific outputs.

Mastering polynomial expressions enables you to tackle a broad range of mathematical problems and deeper calculus concepts, like differentiation and integration, where polynomials are frequently used.