91Ó°ÊÓ

When we talk about the binomial theorem, we are diving into a fundamental concept useful for expanding expressions that are raised to a power. The binomial theorem provides a way to expand expressions of the form \((a + b)^n\), where \(n\) is any positive integer. This is particularly useful in calculus and algebra for simplifying expressions before solving them.In this specific exercise, we needed to expand \((5+h)^2\). The formula for expanding a binomial square is:

• \((a+b)^2 = a^2 + 2ab + b^2\)

Applying the formula, we substitute \(a = 5\) and \(b = h\) to get:

• \(5^2 + 2 \cdot 5 \cdot h + h^2 = 25 + 10h + h^2\)

This expansion is vital as it transforms the original expression \((5+h)^2 - 5^2\) into a simpler form, making it easier to simplify and evaluate later in the process.

Simplifying expressions is an important skill in algebra and calculus. It helps in reducing complex expressions to simpler forms, which are easier to work with. In our step-by-step solution, after expanding the binomial, the expression we were left with was \(25 + 10h + h^2 - 25\).By subtracting \(25\) from the expression, we get:

• \(10h + h^2\)

The next step in simplifying is factoring out common terms. Here, \(h\) can be factored out from the expression:

• \(h(10 + h)\)

This action of factoring is crucial as it allows us to cancel terms later on when evaluating limits. Simplifying expressions involves breaking them down to their smallest and most manageable components, which makes further calculations easier and less prone to errors.

The concept of limits is central to calculus and is used to determine the value that a function approaches as the input approaches a certain point. In this exercise, the aim was to find the limit of the expression as \( h \) tends to zero.Initially, the expression posed was:

• \(\lim_{h \to 0} \frac{(5+h)^2 - 5^2}{h}\)

After simplifying the expression, we reached a stage where we had:

• \(\lim_{h \to 0} \frac{h(10 + h)}{h}\)

By canceling \(h\) from the numerator and the denominator, the expression simplifies to:

• \(\lim_{h \to 0} (10 + h)\)

Finally, by substituting \(h = 0\), we find that the limit is \(10\). This final step shows how limits can be used to find precise values that a function is heading towards as its variable approaches a particular point. It requires simplifying the expression first so that a direct substitution can be made, ensuring an accurate evaluation of the limit.