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Chapter 2: Problem 47

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form $$\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ occur frequently in calculus. In Exercises \(43-48,\) evaluate this limit for the given value of \(x\) and function \(f\). $$ f(x)=\sqrt{x}, \quad x=7 $$

Short Answer

Expert verified
The limit is \( \frac{\sqrt{7}}{14} \).

Step by step solution

01

Understand the Problem

We need to find the limit \( \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \) for the given function \( f(x) = \sqrt{x} \) at \( x = 7 \). This limit is the derivative of the function at \( x = 7 \).
02

Substitute the Function into the Limit Expression

Substitute \( f(x) = \sqrt{x} \) into the limit expression:\[ \lim _{h \rightarrow 0} \frac{\sqrt{7+h}-\sqrt{7}}{h} \].
03

Apply the Conjugate to Simplify the Expression

Multiply and divide the expression by the conjugate of the numerator:\[ \lim_{h \rightarrow 0} \frac{(\sqrt{7+h} - \sqrt{7})(\sqrt{7+h} + \sqrt{7})}{h(\sqrt{7+h} + \sqrt{7})} \].
04

Simplify Using a Difference of Squares

The numerator simplifies to:\((\sqrt{7+h})^2 - (\sqrt{7})^2 = 7+h - 7 = h\).Thus, the limit becomes:\[ \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{7+h} + \sqrt{7})} \].
05

Cancel out \(h\) and Evaluate the Limit

After cancelling \( h \) from the numerator and denominator, we get:\[ \lim_{h \rightarrow 0} \frac{1}{\sqrt{7+h} + \sqrt{7}} \].Evaluate the limit as \( h \rightarrow 0 \), which gives:\[ \frac{1}{\sqrt{7} + \sqrt{7}} = \frac{1}{2\sqrt{7}} \].
06

Simplify Numerical Expression

Simplify the expression: \[ \frac{1}{2\sqrt{7}} = \frac{\sqrt{7}}{14} \] by multiplying the numerator and the denominator by \( \sqrt{7} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limits
In calculus, limits are fundamental in understanding how functions behave as they approach specific points. When dealing with functions like \( f(x) = \sqrt{x} \) and evaluating \( \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \), you essentially look at how a function changes as the input increment \( h \) becomes infinitesimally small. This expression is crucial in determining the slope of the tangent line at any particular point which is vital for finding derivatives.

  • Intuitive Understanding: Think of a limit as observing what happens when you get increasingly closer to a specific point. It's like zooming in on a small section of the function graph.
  • Application: Limits help in understanding continuity, defining derivatives, and evaluating the behavior of functions near singular points.
  • Common Pitfalls: Calculus students often confuse limits with function values. Remember, limits concern approaches, not necessarily actual values.
By understanding limits, you build the foundation for grasping deeper calculus concepts, like derivatives, which describe the rate of change of a function.
Derivatives
In calculus, derivatives are the tools we use to determine how a function changes at any given point. A derivative essentially measures the slope of the tangent line to the curve of a function.For our function \( f(x) = \sqrt{x} \), finding \( \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \) calculates \( f'(x) \), the derivative of \( f \) at point \( x \).

  • Rate of Change: Derivatives tell us how fast something is changing - in mathematical functions, physical motions, or real-world processes.
  • Tangents and Normals: The derivative gives the slope of the tangent line. This is crucial in fields like engineering, physics, and economics.
  • Calculation Techniques: Differentiation rules such as the power rule, product rule, and quotient rule make finding derivatives systematic and straightforward.
Understanding derivatives extends beyond simple calculations. It helps in recognizing how amounts change, optimizing functions, and solving problems involving motion and change.
Calculus Concepts
Calculus is divided into two main branches: differential calculus and integral calculus. Key to these is understanding how quantities change and accumulate.For example, by finding the derivative of \( f(x) = \sqrt{x} \), we can understand how the function behaves around \( x = 7 \).

  • Differential Calculus: Focuses on how things change. Distinctly involves the concepts of limits and derivatives to describe rates of change.
  • Integral Calculus: While not the focus here, it complements derivatives by accumulating quantities. It calculates areas under curves, among other things.
  • The Unifying Principle: The fundamental theorem of calculus creates a bridge between derivatives and integrals, showing how differentiation and integration are inverse processes.
By mastering these calculus concepts, students can fully grasp the depth of mathematics as it applies to natural phenomena, engineering, economics, and beyond. The insights provided by calculus pave the way for solving complex problem scenarios.

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Most popular questions from this chapter

In Exercises \(19-22,\) find the slope of the curve at the point indicated. $$ y=1-x^{2}, \quad x=2 $$

Define \(g(4)\) in a way that extends \(g(x)=\left(x^{2}-16\right) /\) \(\left(x^{2}-3 x-4\right)\) to be continuous at \(x=4 .\)

If the product function \(h(x)=f(x) \cdot g(x)\) is continuous at \(x=0\) , must \(f(x)\) and \(g(x)\) be continuous at \(x=0 ?\) Give reasons for your answer.

Sometimes a change of variable can change an unfamiliar expression into one whose limit we know how to find. For example, $$ \begin{aligned} \lim _{x \rightarrow \infty} \sin \frac{1}{x} &=\lim _{\theta \rightarrow 0^{+}} \sin \theta \\ &=0 \end{aligned} $$ This suggests a creative way to "see" limits at infinity. Describe the procedure and use it to picture and determine limits in Exercises \(79-84 .\) $$ \lim _{x \rightarrow \pm \infty}\left(3+\frac{2}{x}\right)\left(\cos \frac{1}{x}\right) $$

Roots of a cubic Show that the equation \(x^{3}-15 x+1=0\) has three solutions in the interval \([-4,4] .\)
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