Question

Other techniques 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 is $\frac{-1}{25}$.

Step-by-step solution

Find a common denominator and combine the fractions

To find a common denominator and combine the fractions, we can multiply and divide the first fraction by 5 and the second fraction by 5 + h. Then, we can subtract the numerators and combine the denominators. $$\lim _{h \rightarrow 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h} = \lim _{h \rightarrow 0} \frac{\frac{5}{5(5+h)}-\frac{5+h}{5(5+h)}}{h}$$

Combine the fractions and simplify the numerator

Now, we will subtract the numerators of the fractions in the limit expression and simplify the resulting fraction. $$\lim _{h \rightarrow 0} \frac{\frac{5-(5+h)}{5(5+h)}}{h}$$ $$\lim _{h \rightarrow 0} \frac{\frac{-h}{5(5+h)}}{h}$$

Cancel h in the numerator and denominator

At this point, we can cancel h's in the numerator and denominator of our expression. $$\lim _{h \rightarrow 0} \frac{-h}{5(5+h)} \cdot \frac{1}{h} = \lim _{h \rightarrow 0} \frac{-1}{5(5+h)}$$

Evaluate the limit

Now, we can evaluate the limit as h approaches 0. $$\lim _{h \rightarrow 0} \frac{-1}{5(5+h)} = \frac{-1}{5(5+0)} = \frac{-1}{25}$$ So the limit is, $$\lim _{h \rightarrow 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h} = \frac{-1}{25}$$

Key concepts

Understanding Derivatives

In calculus, a derivative represents the rate at which a function is changing at any given point. Essentially, it tells us how a function goes up or down with each small step to the right on the x-axis. In our exercise, we were working with a limit that leads to finding the derivative. This limit form is known as the difference quotient, which is presented as: \[ \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \] For this problem, our function is \(f(x) = \frac{1}{x}\) evaluated at \(x = 5\), and we need to find how the function's value changes as \(h\) gets closer to zero. By applying the steps correctly, we find that the function changes at a rate of \(\frac{-1}{25}\) when \(x = 5\). Understanding this concept is crucial, as derivatives are foundational in studying how functions behave, optimize processes, and even solving real-world problems.

Limit Evaluation Techniques

Limit evaluation involves finding the value that a function approaches as the input approaches a particular point. It’s important in calculus because it often helps us find derivatives, like in this exercise. Here, we evaluated the \( \lim_{h \rightarrow 0} \frac{\frac{1}{5+h} - \frac{1}{5}}{h} \). The approach involves simplifying the expression by :

• Identifying a common denominator for the fractions in the numerator. In our case, it involved multiplying to reach \(5(5+h)\).
• Simplifying the resulting expression, in our example, combining and cancelling terms like \(h\).
• After canceling, simplifying to \(-\frac{1}{5(5+h)}\), and finally replacing \(h\) with 0 to solve \(-\frac{1}{25}\).

This shows how valuable limit evaluation is in calculus; it helps bridge the gap from algebra to calcualtus by finding exact values as approaches are refined to zero!

Simplifying Fractions in Limits

Fraction simplification can be crucial when dealing with limits, especially when working towards derivatives. This exercise involves the subtraction of two fractional values, which requires combining them over a shared denominator. The initial fractions were \(\frac{1}{5+h}\) and \(\frac{1}{5}\). To subtract these, we found a common denominator, \(5(5+h)\), and rewrote each fraction. This process is:

• Multiply the first fraction by \(5\) (numerator and denominator) to make it \(\frac{5}{5(5+h)}\).
• Multiply the second fraction by \(5+h\) to make it \(\frac{5+h}{5(5+h)}\).
• Subtracting these fractions simplifies them to \(\frac{-h}{5(5+h)}\).

Finally, the common factor \(h\) is canceled, giving an expression that is much simpler to evaluate the limit with further refinement. Simplifying fractions helps tailor complex expressions into ones that can be easily analyzed and helps focus on the core behavior of the functions involved.