Problem 21
Question
find the indicated limit or state that it does not exist. In many cases, you will want to do some algebra before trying to evaluate the limit. $$ \lim _{u \rightarrow-2} \frac{u^{2}-u x+2 u-2 x}{u^{2}-u-6} $$
Step-by-Step Solution
Verified Answer
The limit is \( \frac{-2-x}{-5} \).
1Step 1: Analyze the Expression
Examine the given limit \( \lim _{u \rightarrow -2} \frac{u^{2} - u x + 2 u - 2 x}{u^{2} - u - 6} \). We notice that directly substituting \( u = -2 \) into the expression will result in a \( \frac{0}{0} \) indeterminate form, so we must perform algebraic manipulation to simplify the expression.
2Step 2: Factor the Denominator
The denominator is \( u^{2} - u - 6 \). We factor it to find the roots: \[(u - 3)(u + 2)\]This cancels out at the point \( u = -2 \), suggesting there might be a common factor in the numerator.
3Step 3: Simplify the Numerator
Rewrite the numerator to identify common factors: \( u^{2} - u x + 2 u - 2 x \). Group the terms: \( (u^{2} + 2u) - (ux + 2x) \). Factor both groups:\[u(u + 2) - x(u + 2) = (u + 2)(u - x)\]
4Step 4: Cancel Common Factors
Now the expression becomes: \[ \lim_{u \rightarrow -2} \frac{(u + 2)(u - x)}{(u - 3)(u + 2)} \]. Cancel the common factor \((u + 2)\) from the numerator and the denominator.
5Step 5: Evaluate the Simplified Limit
Substitute \( u = -2 \) in the remaining expression: \[ \lim_{u \rightarrow -2} \frac{u - x}{u - 3} = \frac{-2 - x}{-2 - 3} = \frac{-2 - x}{-5} \]. This result is simplified.
Key Concepts
Algebraic ManipulationIndeterminate FormsFactoring PolynomialsLimit Evaluation Techniques
Algebraic Manipulation
Algebraic manipulation is a crucial technique in calculus, especially when dealing with limit problems. In many cases, substitution might first lead to an indeterminate form such as \( \frac{0}{0} \). This requires us to manipulate the expression algebraically to simplify it and avoid problematic forms.
- Start by rearranging terms or grouping them to make them easier to factor.
- Identify common patterns or factors that can cancel out to simplify complex expressions.
Indeterminate Forms
Indeterminate forms, such as \( \frac{0}{0} \), occur in calculus when substitution into a limit results in an undefined expression. These forms signal that the limit needs further evaluation since it cannot be directly calculated from the original equation.
- These forms raise the need for additional algebraic manipulation or other methods to find a defined result.
- They often require recognizing special patterns or applying calculus techniques to resolve the expression into a determinate form.
Factoring Polynomials
Factoring polynomials is essential when simplifying expressions in calculus, especially for limits. It is the process of rewriting a polynomial as a product of its factors, making simplification easier.
- This technique is beneficial for identifying common factors in both the numerator and denominator of an expression.
- By successfully factoring, you can often cancel these common factors, which significantly simplifies the calculations involved.
- Recognizing standard factoring patterns, like difference of squares or trinomial factoring, helps in performing these actions quickly and accurately.
Limit Evaluation Techniques
Limit evaluation techniques are methods used to find the limit of an expression as the variable approaches a certain value. These techniques become necessary when direct substitution is not suitable due to indeterminate forms.
- Factoring: Simplifies the expression by eliminating common factors, as shown in the given exercise.
- Substitution: After simplification, direct substitution often provides the limit.
- L'Hôpital's Rule: Used for \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) forms by differentiating the numerator and denominator separately.
- Simplification Techniques: May involve expanding, combining fractions, or performing algebraic manipulations to reach a solvable form.
Other exercises in this chapter
Problem 20
Find the limits. $$ \lim _{x \rightarrow \infty} \frac{\sqrt{2 x+1}}{x+4} $$
View solution Problem 21
The given function is not defined at a certain point. How should it be defined in order to make it continuous at that point? $$ H(t)=\frac{\sqrt{t}-1}{t-1} $$
View solution Problem 21
Give an \(\varepsilon-\delta\) proof of each limit fact. $$ \lim _{x \rightarrow-1}\left(x^{2}-2 x-1\right)=2 $$
View solution Problem 21
, use a calculator to find the indicated limit. Use a graphing calculator to plot the function near the limit point. $$ \lim _{x \rightarrow 0} \frac{(x-\sin x)
View solution