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Understanding limit calculation is essential in evaluating mathematical expressions as they approach a particular value. In many calculus problems, a limit involves calculating the behaviour of a function as the input, such as a variable, nears a specific number. In this exercise, the solution requires finding the limit as \( h \) approaches zero of the simplified expression \[\lim _{h \rightarrow 0} -5(h+2)\]Here is how it is done in a simple manner:

• Substitute the value that \( h \) approaches into the expression. This means replacing \( h \) with \( 0 \) in the given expression: \(-5(h + 2)\).
• The new expression becomes \(-5(0 + 2)\), which simplifies clearly to \(-10\).

This approach allows us to evaluate the limit straightforwardly. The core idea is based on continuity, meaning that if you plug the value into a simplified expression without causing any undefined situation, the result is the limit. In this example, plugging \( h = 0 \) directly into the expression works smoothly, giving us \(-10\). This simple substitution method avoids complications and quickly leads to the answer.

Algebraic simplification is an essential skill in solving mathematical problems, especially those involving complex expressions.Let's consider this example: Given the expression:\[\frac{5 - 5(h + 1)^2}{h}\]The first step is to expand the polynomial using binomial expansion. The expansion of \((h+1)^2\) results in:

• \( h^2 + 2h + 1 \)

Next, substitute back into the expression to get:\[5 - 5(h^2 + 2h + 1)\]Distribute \(5\) through the binomial terms yielding:\[5 - 5h^2 - 10h - 5\]Combine the Like Terms:

• Notice how \(5 - 5\) cancels out to \(0\).
• The expression becomes \(-5h^2 - 10h\).

Dividing the entire expression by \(h\) now leads us to simplify further, facilitating easy factorization and successive calculations. Simplifying algebraic expressions like these helps to leverage foundational principles that make complex problems manageable.

Factoring polynomials is a fundamental process used in algebra to simplify expressions and solve equations. It involves writing a polynomial as a product of its factors, thus breaking it down into simpler parts.In our exercise, we have an algebraic expression:\[\frac{-5h^2 - 10h}{h}\]We begin by factoring \(h\) from the numerator:

• The terms within the numerator are \(-5h^2\) and \(-10h\).
• Factor \(h\) out: \(-5h(h + 2)\).

The expression simplifies to:\[\frac{-5h(h + 2)}{h}\]Cancelling the \(h\) in both the numerator and the denominator leaves:\[-5(h + 2)\]This demonstrates how factoring polynomials can simplify algebraic expressions and make it easier to apply further mathematical operations, such as finding limits. Recognizing common factors allows us to reduce expressions neatly, highlighting how one variable or term influences the expression's overall form.