Problem 54 Find the limits. \(\lim \frac{... [FREE SOLUTION]

Chapter 2: Problem 54

Find the limits. \(\lim \frac{x}{x^{2}-1}\) as a. \(x \rightarrow 1^{+}\) b. \(x \rightarrow 1^{-}\) c. \(x \rightarrow-1^{+}\) d. \(x \rightarrow-1^{-}\)

Short Answer

Expert verified

a. \(+\infty\), b. \(-\infty\), c. \(0^-\), d. \(0^+\)

Step by step solution

Understand the Function

We are given the function \( f(x) = \frac{x}{x^2 - 1} \). This function has a denominator \( x^2 - 1 \) which can be factored into \((x-1)(x+1)\). The points of interest where the denominator becomes zero are \(x = 1\) and \(x = -1\). It implies potential vertical asymptotes or undefined points at these values.

Limit as x approaches 1 from the right (\(x \to 1^+\))

As \(x\) approaches 1 from the right, \( x \to 1^+\), the denominator \((x-1)\) approaches \(0^+\) and \((x+1)\) approaches \(2\). The overall expression approaches \(+\infty\). Hence, \(\lim_{{x \to 1^+}} \frac{x}{x^2 - 1} = +\infty\).

Limit as x approaches 1 from the left (\(x \to 1^-\))

As \(x\) approaches 1 from the left, \( x \to 1^-\), \((x-1)\) approaches \(0^-\) and \((x+1)\) still approaches \(2\). The overall expression approaches \(-\infty\). Therefore, \(\lim_{{x \to 1^-}} \frac{x}{x^2 - 1} = -\infty\).

Limit as x approaches -1 from the right (\(x \to -1^+\))

As \(x\) approaches -1 from the right, \( x \to -1^+\), \((x+1)\) approaches \(0^+\) and \((x-1)\) approaches \(-2\). The overall expression approaches \(0^-\) or \(0\) from the negative side. Thus, \(\lim_{{x \to -1^+}} \frac{x}{x^2 - 1} = 0^-\).

Limit as x approaches -1 from the left (\(x \to -1^-\))

As \(x\) approaches -1 from the left, \( x \to -1^-\), \((x+1)\) approaches \(0^-\) and \((x-1)\) again approaches \(-2\). The overall expression approaches \(0^+\) or \(0\) from the positive side. Therefore, \(\lim_{{x \to -1^-}} \frac{x}{x^2 - 1} = 0^+\).

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Key Concepts

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

Vertical Asymptotes

Vertical asymptotes occur in a graph where a function approaches infinity or negative infinity as it nears a specific value. They signify places where the function does not exist. In our function \(f(x) = \frac{x}{x^2 - 1}\), the denominator \(x^2 - 1\) becomes zero at \(x = 1\) and \(x = -1\). This creates vertical asymptotes at these points. To understand this better:

As \(x\) gets very close to these values, the function will shoot up to positive or negative infinity.
Vertical asymptotes indicate undefined points but also suggest limit behavior near these values.

So, for limits approaching \(x = 1\) or \(x = -1\), we need to employ one-sided limits to analyze how the function behaves just before it reaches these points.

One-Sided Limits

One-sided limits help examine the behavior of a function as \(x\) approaches a particular value from one direction only. They provide more insight, especially around vertical asymptotes.

For \(x \to 1^+\), the denominator \((x - 1)\) tends to \(0^+\) while \((x + 1)\) heads towards \(2\). Hence, the function approaches \(+\infty\).
For \(x \to 1^-\), \((x - 1)\) approaches \(0^-\), making the function \(-\infty\).

For \(x \to -1^+\), \((x + 1)\) approaches \(0^+\) while \((x - 1)\) goes towards \(-2\). The function thus tends to \(0^-\).
For \(x \to -1^-\), \((x + 1)\) trends \(0^-\) and \((x - 1)\) remains \(-2\), directing the function towards \(0^+\).

These insights are vital for understanding precise behavior near critical points on the graph.

Factoring Polynomials

Factoring polynomials simplifies functions to reveal more about their behavior. In our example, the polynomial in the denominator \(x^2 - 1\) was factored as \((x-1)(x+1)\). This crucial step helps identify vertical asymptotes and allows for the computation of limits. Here鈥檚 why factoring is key:

By observing factorization, we identify points where the function is undefined due to division by zero.
It enables the application of limits, especially determining one-sided limits around tricky spots.
Factoring polynomials reduces complexity, often turning polynomials into simpler linear functions or binomials to analyze.

Always look out for opportunities to factor expressions within complex fractions, as it can provide clarity and simplicity in handling limits and other calculus operations.

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91影视

Find the limits. \(\lim \frac{x}{x^{2}-1}\) as a. \(x \rightarrow 1^{+}\) b. \(x \rightarrow 1^{-}\) c. \(x \rightarrow-1^{+}\) d. \(x \rightarrow-1^{-}\)

Short Answer

Step by step solution

Understand the Function

Limit as x approaches 1 from the right (\(x \to 1^+\))

Limit as x approaches 1 from the left (\(x \to 1^-\))

Limit as x approaches -1 from the right (\(x \to -1^+\))

Limit as x approaches -1 from the left (\(x \to -1^-\))

Key Concepts

Vertical Asymptotes

One-Sided Limits

Factoring Polynomials

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