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Finding a Limit In Exercises \(47-62,\) find the limit. $$ \lim _{x \rightarrow 0} \frac{x}{x^{2}-x} $$

Short Answer

Expert verified
The limit of the given function as \(x\) approaches 0 is \(-1\).

Step by step solution

01

Factor the Denominator

Factor the denominator to simplify the faction. This can be done by recognizing that both terms in the denominator share an \(x\) factor. The factored form of \(x^{2} - x\) is \(x(x - 1)\).
02

Simplify the Fraction

The fraction becomes simplified when we divide \(x\) in the numerator by the common \(x\) in the denominator. The simplified form becomes \(\frac{1}{x - 1}\).
03

Evaluate the Limit

Substitute \(x = 0\) into the simplified function, \(\frac{1}{x - 1}\), to obtain the answer \(-1\).

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Key Concepts

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

Understanding the Limit of a Function
In calculus, the concept of a limit is fundamental for analyzing how functions behave as they approach a specific point. At its core, a limit answers the question: What value is the function getting closer and closer to, as the input (or 'x' value) approaches a certain number? This is crucial in understanding how functions grow near points where they may not be clearly defined, such as when a function potentially divides by zero.

For example, the limit as \(x\) approaches 0 for the function \(\frac{x}{x^2 - x}\) seeks the value the function approaches as \(x\) gets closer to 0. By finding the limit, we can understand the behavior of the function around that point which could be inconspicuous by direct substitution due to undefined expressions or indeterminate forms.
Factoring Polynomials
The procedure of rewriting a polynomial as a product of its factors is called factoring. Factoring is not just a mathematical trick; it is a powerful tool that simplifies expressions and solves equations. When working with limits, factoring can be especially useful because it can eliminate common factors in the numerator and denominator, which can clear up indeterminate forms like 0/0.

In our exercise, the denominator \(x^2 - x\) can be factored into \(x(x - 1)\). This is possible as both terms share an \(x\) factor, showing the importance of recognizing common factors to simplify expressions before determining their limits.
Simplifying Fractions in Calculus
Simplifying fractions is a key step in finding limits involving rational functions. This process often involves reducing the fraction to its lowest terms, which can eliminate potential undefined values and indeterminacies. By doing this, we can accurately evaluate the limit of a function as the variable approaches a particular value.

In this case, once we have factored the denominator to \(x(x - 1)\), we notice the common \(x\) in both the numerator and denominator. This allows us to cancel out the \(x\) and simplifies our fraction to \(\frac{1}{x - 1}\), which can then be directly evaluated for the limit as \(x\) approaches 0.
The Substitution in Limits
Once we have simplified the function, we can often use direct substitution to find the limit, provided that the substitution does not result in an undefined expression. If the function is continuous at the point to which the limit approaches, direct substitution will yield the limit's value.

In the given exercise, after simplifying the fraction and removing any common factors, we substitute \(x = 0\) into the simplified function \(\frac{1}{x - 1}\) to find that the limit is \(-1\). This final step represents applying the limit concept to the simplified expression, leading to the understanding of the function's behavior at the specified point.

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Most popular questions from this chapter

Finding a Limit In Exercises \(7-26\) , find the limit (if it exists). If it does not exist, explain why. $$ \lim _{x \rightarrow 1^{+}} f(x), \text { where } f(x)=\left\\{\begin{array}{ll}{x,} & {x \leq 1} \\ {1-x,} & {x>1}\end{array}\right. $$

Writing In Exercises \(87-90\) , explain why the function has a zero in the given interval. $$ f(x)=x^{2}-2-\cos x \quad[0, \pi] $$

Telephone Charges A long distance phone service charges \(\$ 0.40\) for the first 10 minutes and \(\$ 0.05\) for each additional minute or fraction thereof. Use the greatest integer function to write the cost \(C\) of a call in terms of time \(t\) (in minutes). Sketch the graph of this function and discuss its continuity.

Using the Intermediate Value Theorem In Exercises \(91-94,\) use the Intermediate Value Theorem and a graphing utility to approximate the zero of the function in the interval \([0,1] .\) Repeatedly "zoom in" on the graph of the function to approximate the zero accurate to two decimal places. Use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places. $$ h(\theta)=\tan \theta+3 \theta-4 $$

Using the Intermediate Value Theorem In Exercises \(91-94,\) use the Intermediate Value Theorem and a graphing utility to approximate the zero of the function in the interval \([0,1] .\) Repeatedly "zoom in" on the graph of the function to approximate the zero accurate to two decimal places. Use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places. $$ f(x)=x^{4}-x^{2}+3 x-1 $$

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