Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{h \rightarrow 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h}$$

Short answer

Question: Evaluate the limit of the given expression: $$\lim _{h \rightarrow 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h}$$ Answer: The limit of the given expression is \(\frac{-1}{25}\).

Step-by-step solution

Rewrite the expression

We need to simplify the given expression into a more manageable form. Let's rewrite the expression as: $$\lim _{h \rightarrow 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h} = \lim_{h \rightarrow 0} \frac{1}{h} \left(\frac{1}{5+h}-\frac{1}{5}\right)$$

Finding the common denominator

In order to simplify the expression in the parenthesis, we need to find the common denominator: $$\lim_{h \rightarrow 0} \frac{1}{h} \left(\frac{5 - (5 + h)}{5(5+h)}\right)$$

Simplifying the expression

Now, simplify the expression in the parentheses: $$\lim_{h \rightarrow 0} \frac{1}{h} \left(\frac{-h}{5(5+h)}\right)$$

Cancel out the h term

Notice that the \(h\) term can be canceled out in the numerator and denominator: $$\lim_{h \rightarrow 0} \left(\frac{-1}{5(5+h)}\right)$$

Evaluate the limit

Now, we can plug \(h=0\) into the expression to evaluate the limit: $$\lim_{h \rightarrow 0} \left(\frac{-1}{5(5+h)}\right) = \frac{-1}{5(5+0)} = \frac{-1}{25}$$ The limit of the given expression is \(\frac{-1}{25}\).

Key concepts

Algebraic Simplification

Algebraic simplification is crucial in limit calculus as it makes complex expressions easier to handle. In the given exercise, we simplify the expression to find the limit. Initially, the expression is \( \lim _{h \rightarrow 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h} \). This can seem quite complex at first glance.
To simplify it, we first rewrite the expression by setting it up in a more recognizable form: \[\lim_{h \rightarrow 0} \frac{1}{h} \left(\frac{1}{5+h}-\frac{1}{5}\right)\]. Next, we aim to consolidate the terms within the parentheses by converting them into a single fraction.
This involves finding a common denominator, which, in this case, is \(5(5+h)\). Simplifying algebraic expressions often requires such operations to make calculations more straightforward.

Limit Evaluation

When we talk about limit evaluation, we refer to determining what value a function approaches as the input reaches a particular point. For this exercise, our goal is to evaluate the limit \(\lim_{h \rightarrow 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h}\). Here, \(h\) approaches zero, which often means the expression's initial form doesn't allow direct substitution.
Limiting evaluations generally involve simplifying troublesome terms so that they don't make calculations impossible. In this case, the simplified expression \(\lim_{h \rightarrow 0} \left(\frac{-1}{5(5+h)}\right)\) lets us straightforwardly substitute \(h = 0\) to get \(-1/25\). This indicates the function's value as \(h\) tends to zero.
The steps leading to this are incremental. It involves algebraic simplification followed by careful substitution.

Rational Expressions

Rational expressions are fractions where both the numerator and denominator are polynomials. In this exercise, the given expression is rational because it consists of fractions involving polynomials in \(h\). Handling these requires some careful algebra.
We started with the rational expression \(\frac{\frac{1}{5+h}-\frac{1}{5}}{h}\). To manage it effectively, finding a common denominator within the parentheses was essential. We converted the subtraction within the numerator into a single fraction. This step was necessary to simplify the expression to something workable where certain terms would cancel out.
Rational expressions are typically challenging due to singularities and undefined points like the zero in the denominator. By simplifying the expression and evaluating limits, we can resolve these issues and find meaningful values where they initially appear problematic.