Question

Evaluate the following limits, where a and \(b\) are fixed real numbers. \(\lim _{h \rightarrow 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h}\)

Short answer

Answer: The value of the limit as \(h\) approaches 0 for the given expression is \(\frac{-1}{25}\).

Step-by-step solution

Simplify the expression in the limit

To begin simplifying, we will first find a common denominator for the two fractions in the numerator: \[\frac{1}{5+h} - \frac{1}{5} = \frac{5 - (5+h)}{5(5+h)} = \frac{-h}{5(5+h)}\] Now, substitute this back into the original expression: \[\frac{\frac{1}{5+h}-\frac{1}{5}}{h} = \frac{\frac{-h}{5(5+h)}}{h}\]

Cancel out the "h" term

In this step, we're going to cancel out the h term in the numerator and the denominator: \[\frac{\frac{-h}{5(5+h)}}{h} = \frac{-1}{5(5+h)}\]

Find the limit as h approaches 0

Now that we have simplified the expression, we can find the limit as h approaches 0: \[\lim_{h \rightarrow 0} \frac{-1}{5(5+h)} = \frac{-1}{5(5+0)} = \frac{-1}{25}\] The limit of the given expression as \(h\) approaches 0 is \(\frac{-1}{25}\).

Key concepts

Limits of a Function

Understanding the concept of a limit is crucial in calculus. In layman's terms, the limit of a function describes what happens to the function's value as the input approaches a certain point. For instance, with the exercise provided, we are interested in finding the value that the function \(\frac{1}{5+h}-\frac{1}{5}\) gets closer and closer to, as \(h\) approaches zero.

It's essential to note that the limit does not need to be the function's actual value at that point; it's the value the function is tending towards. This concept serves as the foundation for understanding continuous and discontinuous points in a function and sets the groundwork for derivatives, which measure the rate of change of functions.

Limit Properties

When dealing with limits, there are several properties that make computing them more systematic. These include the sum, product, quotient, and power rules. By using these properties, we can simplify complex expressions into more manageable parts that can be easily evaluated.

For example, when the limits of two functions exist separately, the limit of their sum equals the sum of their limits. In the exercise above, initially, you may have two separate limits, but they are combined using these algebraic properties to form a simpler expression. This use of limit properties helps to find the limit of a function as the input approaches a particular value.

Algebraic Manipulation of Limits

Solving limits often requires a series of algebraic manipulations to simplify the expression to a point where the limit can be directly evaluated. This could involve factoring, expanding, rationalizing, or canceling terms. The goal is to eliminate the part that's causing the indeterminate form, such as a zero in the denominator or a difference of zeroes.

In the provided exercise, we executed algebraic manipulation by finding a common denominator and then canceling out the \(h\) term. Without these steps, directly substituting \(h = 0\) into the original expression would result in an indeterminate form of \(\frac{0}{0}\), which is undefined. Such manipulation ensures that limits can be calculated accurately and efficiently.

Simplifying Complex Fractions

Complex fractions can be intimidating at first glance, with numerators and denominators within a larger fraction. Simplifying such expressions is a key skill in evaluating limits. The exercise requires you to simplify a complex fraction wherein the numerator itself contains a fraction. This is done by finding common denominators, combining terms, and reducing wherever possible.

One effective simplification strategy is to multiply the numerator and denominator by the least common denominator (LCD) of all the fractional components. In this case, \(h\) was common to both the numerator and denominator, which allowed us to cancel it out and simplify the expression before taking the limit. Mastery of simplifying complex fractions leads to an easier time with limit computations and other areas of calculus.