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Chapter 2: Problem 10

Compute the limits. If a limit does not exist, explain why. $$ \lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1} \Rightarrow $$

Short Answer

Expert verified
The limit is 2.

Step by step solution

01

Identify the Limit Expression

The expression given is \( \lim_{x \rightarrow 1} \frac{x^2 - 1}{x - 1} \). First, recognize that \( x^2 - 1 \) can be factored using the difference of squares.
02

Factor the Numerator

Factor the expression \( x^2 - 1 \) as \((x - 1)(x + 1)\). So, the expression becomes \( \frac{(x-1)(x+1)}{x-1} \).
03

Simplify the Expression

Cancel out the common factor \((x - 1)\) in the numerator and denominator. This simplifies the expression to \( x + 1 \).
04

Evaluate the Simplified Expression at the Limit Point

Substitute \( x = 1 \) in the simplified expression \( x + 1 \), resulting in \( 1 + 1 = 2 \).
05

State the Limit

Since the expression \( x + 1 \) is continuous at \( x = 1 \), the limit as \( x \rightarrow 1 \) is \( 2 \). Therefore, \( \lim_{x \rightarrow 1} \frac{x^2 - 1}{x - 1} = 2 \).

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Key Concepts

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

Difference of Squares
The difference of squares is a powerful algebraic tool used to simplify expressions where a square number is subtracted from another square. Recognizing this pattern is crucial in calculous and algebra. This formula is given as:
\[(a^2 - b^2) = (a - b)(a + b)\]
In our exercise, we dealt with the expression \(x^2 - 1\). By noticing it fits the form of \(a^2 - b^2\), where \(a = x\) and \(b = 1\), we factor it into \((x - 1)(x + 1)\).
  • This method helps in breaking down complex expressions into simpler, manageable parts.
  • It is particularly useful for solving limits, equations, or even some quadratic expressions.
Recognizing the difference of squares early on makes calculations smoother and helps avoid potential errors.
Simplifying Expressions
Simplifying expressions is a fundamental skill in calculus. It involves reducing an expression to its simplest form while maintaining its value. This process makes solving problems, like limits, more efficient. In our case, after factoring the expression as
\[\frac{(x-1)(x+1)}{x-1}\]
we simplify by canceling out the common factor \((x-1)\), leaving us with \(x + 1\). Here's how to approach simplifying expressions effectively:
  • Look for common factors in the numerator and the denominator to reduce fractions.
  • Identify special patterns (like the difference of squares) to factor quickly.
  • Always simplify step-by-step to avoid losing components accidentally.
Simplifying expressions not only makes equations easier to manage but also prevents errors during further calculations, such as evaluating limits.
Continuous Functions
Understanding when a function is continuous is essential in calculus, especially when dealing with limits. A function is continuous at a point if the limit as it approaches that point equals the function鈥檚 value at that point.
For our exercise, after simplifying the expression to \(x + 1\), we find it is continuous at \(x = 1\). To check continuity:
  • The function must be defined at the point.
  • The limit of the function as \(x\) approaches the point must exist.
  • The limit value must equal the function's value at that point.
In our problem, since all these conditions are met for \(x + 1\) at \(x = 1\), it is continuous there, allowing us to find that the limit is 2. Recognizing continuous functions helps ensure accurate calculations and validate that limit solutions are correct.

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Most popular questions from this chapter

\(\lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)\) (Hint: Use the fact that \(|\sin a|<1\) for any real number a. You should probably use the definition of a limit here.) \(\Rightarrow\)

Sketch the unit circle. Discuss the behavior of the slope of the tangent line at various angles around the circle. Which trigonometric function gives the slope of the tangent line at an angle \(\theta\) ? Why? Hint: think in terms of ratios of sides of triangles.

Compute the limits. If a limit does not exist, explain why. $$ \lim _{x \rightarrow 4} 3 x^{3}-5 x \Rightarrow $$

Let \(y=f(t)=t^{2},\) where \(t\) is the time in seconds and \(y\) is the distance in meters that an object falls on a certain airless planet. Draw a graph of this function between \(t=0\) and \(t=3\). Make a table of the average speed of the falling object between (a) \(2 \mathrm{sec}\) and \(3 \mathrm{sec}\), (b) 2 sec and 2.1 sec \(,\) (c) 2 sec and 2.01 sec, (d) 2 sec and 2.001 sec. Then use algebra to find a simple formula for the average speed between time 2 and time \(2+\Delta t\). (If you substitute \(\Delta t=1,0.1,0.01,0.001\) in this formula you should again get the answers to parts \((\mathrm{a})-(\mathrm{d}) .)\) Next, in your formula for average speed (which should be in simplified form) determine what happens as \(\Delta t\) approaches zero. This is the instantaneous speed. Finally, in your graph of \(y=t^{2}\) draw the straight line through the point (2,4) whose slope is the instantaneous velocity you just computed; it should of course be the tangent line. \(\Rightarrow\)

Find the derivative of \(y=f(x)=\sqrt{169-x^{2}} \cdot \Rightarrow\)
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