Question

Let \(f(x)=x^{3}+x\) Evaluate \(\lim _{s \rightarrow 0} \frac{f(x+s)-f(x)}{s}\).

Short answer

The limit is \( 3x^2 + 1 \).

Step-by-step solution

Understand the Problem

The expression given, \( \lim_{s \rightarrow 0} \frac{f(x+s)-f(x)}{s} \), is the limit definition of the derivative of \( f(x) \) with respect to \( x \). We'll calculate the derivative of \( f(x) \) using this expression.

Express f(x+s)

To calculate \( f(x+s) \), substitute \( x+s \) into the function \( f(x) = x^3 + x \). This gives \( f(x+s) = (x+s)^3 + (x+s) \).

Expand (x+s)^3

Expand the expression \( (x+s)^3 \) using the binomial theorem: \[ (x+s)^3 = x^3 + 3x^2s + 3xs^2 + s^3 \].

Substitute and Simplify

Substitute the expanded \( (x+s)^3 \) back into \( f(x+s) \). This gives: \[ f(x+s) = x^3 + 3x^2s + 3xs^2 + s^3 + x + s \]. Next, simplify \( f(x+s) - f(x) \):\[ f(x+s) - f(x) = (x^3 + 3x^2s + 3xs^2 + s^3 + x + s) - (x^3 + x) \]\[ = 3x^2s + 3xs^2 + s^3 + s \].

Divide by s

Now divide the expression \( f(x+s) - f(x) \) by \( s \): \[ \frac{3x^2s + 3xs^2 + s^3 + s}{s} = 3x^2 + 3xs + s^2 + 1 \].

Take the Limit as s Approaches Zero

Evaluate the limit as \( s \rightarrow 0 \): \[ \lim_{s \rightarrow 0} (3x^2 + 3xs + s^2 + 1) = 3x^2 + 0 + 0 + 1 \]. Thus, the limit is \( 3x^2 + 1 \).

Key concepts

Binomial Theorem

The binomial theorem is a pivotal concept in algebra. It provides a formula to expand expressions of the form \((x+s)^n\). This formula states that:\[(x+s)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} s^k\]where \( \binom{n}{k} \) is a binomial coefficient, also known as "n choose k," and it represents the number of ways to choose \(k\) elements from \(n\) elements.In the exercise, we used the binomial theorem to expand \((x+s)^3\). By applying the formula, we get:

• \(x^3\) from \(k=0\)
• \(3x^2s\) from \(k=1\)
• \(3xs^2\) from \(k=2\)
• \(s^3\) from \(k=3\)

This expansion helps simplify the expression and eventually calculate the derivative by the limit process.

Expanding Polynomials

Expanding polynomials involves breaking down expressions containing variables raised to higher powers. It's an essential skill in algebra since it simplifies complex polynomials and enables us to see individual terms.In this exercise, the polynomial \(f(x+s)\) was expanded using polynomials rules and the binomial theorem. This is to transform expressions like \((x+s)^3\) into simpler terms. Each term of \(s\) in the polynomial \(f(x+s) = (x+s)^3 + (x+s)\) was expanded:

• \((x+s)^3\) was expanded using the binomial theorem, giving us the terms \(x^3 + 3x^2s + 3xs^2 + s^3\).
• \((x+s)\) was directly expanded, providing us with the terms \(x + s\).

Through proper expansion, the expression becomes manageable, facilitating subtraction and further simplification steps in calculus.

Taking Limits

The concept of taking limits is fundamental in calculus, particularly in determining derivatives and areas under curves. The limit addresses what happens as one variable approaches a specific value—often zero in differential calculus.In this exercise, we need to find:\[\lim_{s \rightarrow 0} \frac{f(x+s)-f(x)}{s}\]By simplifying \(f(x+s) - f(x)\) algebraically and dividing by \(s\), we end up with:\[3x^2 + 3xs + s^2 + 1\]Finally, evaluating the limit as \(s\) approaches zero entails substituting \(s = 0\) in the expression:

• \(3x^2\) remains as it depends on \(x\)
• \(3xs\) and \(s^2\) vanish since \(s=0\)
• The constant \(1\) remains

Thus, the limit is \(3x^2 + 1\), which represents the derivative of the original function \(f(x) = x^3 + x\).