Problem 32
Question
Evaluate the limit, if it exists. \( \displaystyle \lim_{h \to 0}\frac{\frac{1}{(x + h)^2} - \frac{1}{x^2}}{h} \)
Step-by-Step Solution
Verified Answer
The limit is \( \frac{-2}{x^3} \).
1Step 1: Recognize the Limit as a Derivative
Notice that the expression resembles the definition of a derivative. Specifically, this limit is the derivative of the function \( f(x) = \frac{1}{x^2} \) at a point \( x \). The general derivative is given by:\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \] Thus, the expression given, \( \lim_{h \to 0} \frac{\frac{1}{(x + h)^2} - \frac{1}{x^2}}{h} \), represents \( f'(x) \).
2Step 2: Find a Common Denominator
Rewriting the expression as a single fraction can simplify it. Here we have:\[ \frac{1}{(x+h)^2} - \frac{1}{x^2} = \frac{x^2 - (x+h)^2}{x^2 (x+h)^2} \]Expanding \((x + h)^2\) gives:\[ x^2 + 2xh + h^2 \]So our expression becomes:\[ \frac{x^2 - x^2 - 2xh - h^2}{x^2 (x+h)^2} = \frac{-2xh - h^2}{x^2 (x+h)^2} \]
3Step 3: Factor out \( h \) from the Numerator
In the expression \( -2xh - h^2 \), factor out an \( h \):\[ -2xh - h^2 = h(-2x - h) \]This gives us:\[ \frac{h(-2x - h)}{x^2 (x + h)^2} \]
4Step 4: Simplify and Cancel \( h \)
Now, simplify the expression by canceling the \( h \) in the numerator and the \( h \) in the denominator:\[ \frac{-2x - h}{x^2 (x + h)^2} \]
5Step 5: Evaluate the Limit as \( h \to 0 \)
Once simplified, evaluate the limit as \( h \to 0 \):\[ \lim_{h \to 0} \frac{-2x - h}{x^2 (x + h)^2} = \frac{-2x}{x^2 \cdot x^2} = \frac{-2x}{x^4} = \frac{-2}{x^3} \]
Key Concepts
DerivativeDifference QuotientRational Functionsh Approaching Zero
Derivative
The concept of a derivative is foundational in calculus. It represents the rate at which a function changes at any given point. In simpler terms, the derivative tells us "how steep" a function is at a specific point.
To find the derivative, we often use the definition involving limits and the difference quotient:
To find the derivative, we often use the definition involving limits and the difference quotient:
- The difference quotient is given by \( \frac{f(x+h) - f(x)}{h} \).
- To find the derivative \( f'(x) \), we evaluate the limit \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).
Difference Quotient
The difference quotient is a crucial part of finding the derivative. It is an expression that represents the average rate of change of the function over a small interval.
- Its formula is \( \frac{f(x+h) - f(x)}{h} \).
- This measures the slope of the secant line connecting two points on the function, \((x, f(x))\) and \((x+h, f(x+h))\).
Rational Functions
Rational functions are quotients of polynomial functions. A typical form is \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. They can often be identified by the presence of variables in the denominator.
- In our example, \( f(x) = \frac{1}{x^2} \) is a rational function.
- The techniques for finding limits or derivatives for these functions involve algebraic manipulation, such as finding common denominators or factoring.
h Approaching Zero
Understanding \( h \) approaching zero is key in limit evaluation. In calculus, we use \( h \) as a small increment to examine how functions behave at a point when the increment becomes infinitesimally small.
- We often use the expression "as \( h \to 0 \)" to describe this process.
- This process helps transition from a discrete to a continuous perspective.
Other exercises in this chapter
Problem 32
Find the limit or show that it does not exist. \( \displaystyle \lim_{x \to \infty} (e^{-x} + 2 \cos 3x) \)
View solution Problem 32
Prove the statement using the \( \varepsilon \), \( \delta \) definition of a limit. \( \displaystyle \lim_{x \to 2} x^3 = 8 \)
View solution Problem 32
Determine the infinite limit. \( \displaystyle \lim_{x \to 5^-}\frac{x+1}{x-5} \)
View solution Problem 33
(a) If \( f(x) = x^4 + 2x \), find \( f'(x) \). (b) Check to see that your answer to part (a) is reasonable by comparing the graphs of \( f \) and \( f' \).
View solution