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Magazine Chapter 3: Problem 11
Find the following limits. (a) \(\lim _{x \rightarrow a} \frac{x^{2}-a^{2}}{x-a}\) (b) \(\lim _{x \rightarrow b} \frac{\sqrt{x}-\sqrt{b}}{x-b}, b>0\) (c) \(\lim _{x \rightarrow a} \frac{x^{3}-a^{3}}{x-a}\)
Short Answer
Expert verified
(a) 2a, (b) \(\frac{1}{2\sqrt{b}}\), (c) 3a^2.
Step by step solution
01
for (a) - Simplify the Expression
We are given the limit \(\frac{x^2 - a^2}{x - a}\). Recognize the numerator as a difference of squares: \(x^2 - a^2 = (x - a)(x + a)\), so the expression simplifies to: \(\frac{(x - a)(x + a)}{x - a}\).
02
for (a) - Cancel Common Factors
Cancel the common factor \(x - a\): \(\frac{(x - a)(x + a)}{x - a} = x + a\).
03
for (a) - Compute the Limit
Now, since the expression is simplified, find the limit as \(x\) approaches \(a\): \(\lim_{x \rightarrow a} (x + a) = a + a = 2a\).
04
for (b) - Rationalize the Numerator
We are given the limit \(\frac{\sqrt{x} - \sqrt{b}}{x - b}\). Multiply the numerator and the denominator by the conjugate of the numerator: \(\frac{\sqrt{x} - \sqrt{b}}{x - b} \cdot \frac{\sqrt{x} + \sqrt{b}}{\sqrt{x} + \sqrt{b}}\).
05
for (b) - Simplify
The numerator becomes a difference of squares: \((\sqrt{x})^2 - (\sqrt{b})^2 = x - b\). The expression simplifies to: \(\frac{x - b}{(x - b)(\sqrt{x} + \sqrt{b})}\).
06
for (b) - Cancel Common Factors
Cancel the common factor \(x - b\): \(\frac{x - b}{(x - b)(\sqrt{x} + \sqrt{b})} = \frac{1}{\sqrt{x} + \sqrt{b}}\).
07
for (b) - Compute the Limit
Find the limit as \(x\) approaches \(b\): \(\lim_{x \rightarrow b} \frac{1}{\sqrt{x} + \sqrt{b}} = \frac{1}{\sqrt{b} + \sqrt{b}} = \frac{1}{2\sqrt{b}}\).
08
for (c) - Factor the Numerator
We are given the limit \(\frac{x^3 - a^3}{x - a}\). Recognize the numerator as a difference of cubes: \(x^3 - a^3 = (x - a)(x^2 + ax + a^2)\).
09
for (c) - Cancel Common Factors
Cancel the common factor \(x - a\): \(\frac{(x - a)(x^2 + ax + a^2)}{x - a} = x^2 + ax + a^2\).
10
for (c) - Compute the Limit
Find the limit as \(x\) approaches \(a\): \(\lim_{x \rightarrow a} (x^2 + ax + a^2) = a^2 + a^2 + a^2 = 3a^2\).
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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 fundamental algebraic concept. It states that for any two real numbers, \(a\) and \(b\), the difference between their squares can be factored as follows: \[x^2 - a^2 = (x - a)(x + a)\].
This is useful in calculus when simplifying expressions before taking limits.
For example, in problem (a) \(\lim _{x \rightarrow a} \frac{x^{2}-a^{2}}{x-a}\), we recognize that \(x^2 - a^2\) is a difference of squares. We can factor it into \( (x - a)(x + a)\).
This allows us to cancel out common factors and simplify the expression, making it easier to find the limit.
This is useful in calculus when simplifying expressions before taking limits.
For example, in problem (a) \(\lim _{x \rightarrow a} \frac{x^{2}-a^{2}}{x-a}\), we recognize that \(x^2 - a^2\) is a difference of squares. We can factor it into \( (x - a)(x + a)\).
This allows us to cancel out common factors and simplify the expression, making it easier to find the limit.
Rationalizing the Numerator
Rationalizing the numerator is a technique used to simplify limits involving square roots. This means converting the numerator from a radical expression to a rational form.
For problem (b) \( \lim _{x \rightarrow b} \frac{\text{鈭歺}-\text{鈭歜}}{x-b}\), we rationalize the numerator by multiplying both the numerator and the denominator by the conjugate of the numerator: \[ \frac{\text{鈭歺}-\text{鈭歜}}{x-b} \cdot \frac{\text{鈭歺} + \text{鈭歜}}{\text{鈭歺} + \text{鈭歜}} \].
The result, when simplified, becomes a difference of squares in the numerator: \[ \frac{x - b}{(x - b)(\text{鈭歺} + \text{鈭歜})}\].
After cancelling the common term \(x - b\), we are left with \[\frac{1}{\text{鈭歺} + \text{鈭歜}} \], making it much easier to find the limit.
For problem (b) \( \lim _{x \rightarrow b} \frac{\text{鈭歺}-\text{鈭歜}}{x-b}\), we rationalize the numerator by multiplying both the numerator and the denominator by the conjugate of the numerator: \[ \frac{\text{鈭歺}-\text{鈭歜}}{x-b} \cdot \frac{\text{鈭歺} + \text{鈭歜}}{\text{鈭歺} + \text{鈭歜}} \].
The result, when simplified, becomes a difference of squares in the numerator: \[ \frac{x - b}{(x - b)(\text{鈭歺} + \text{鈭歜})}\].
After cancelling the common term \(x - b\), we are left with \[\frac{1}{\text{鈭歺} + \text{鈭歜}} \], making it much easier to find the limit.
Difference of Cubes
The difference of cubes is another important algebraic factorization. For any real numbers \(a\) and \(b\), it states: \[ x^3 - a^3 = (x - a)(x^2 + ax + a^2)\].
This is helpful when dealing with polynomial limits.
In problem (c) \(\lim _{x \rightarrow a} \frac{x^{3}-a^{3}}{x-a}\), recognizing \(x^3 - a^3\) as a difference of cubes allows us to factor it: \[\frac{(x - a)(x^2 + ax + a^2)}{x-a} \].
After cancelling the common factor \(x - a\), we are left with the simplified expression \[ x^2 + ax + a^2 \]. By substituting \((x \rightarrow a) \), we can easily compute the limit resulting in \(3a^2\).
This is helpful when dealing with polynomial limits.
In problem (c) \(\lim _{x \rightarrow a} \frac{x^{3}-a^{3}}{x-a}\), recognizing \(x^3 - a^3\) as a difference of cubes allows us to factor it: \[\frac{(x - a)(x^2 + ax + a^2)}{x-a} \].
After cancelling the common factor \(x - a\), we are left with the simplified expression \[ x^2 + ax + a^2 \]. By substituting \((x \rightarrow a) \), we can easily compute the limit resulting in \(3a^2\).
