Question

Find the limit. \(\lim _{x \rightarrow-\infty}\left(\frac{2 x}{x-1}+\frac{3 x}{x+1}\right)\)

Short answer

The limit as x approaches negative infinity of \((\frac{2x}{x - 1} + \frac{3x}{x + 1})\) is 0

Step-by-step solution

Simplify the function

First, the function needs to be simplified. Rational functions can be added together by finding a common denominator. The common denominator in this case is \(x(x - 1)(x + 1)\). This gives: \(\frac{2x(x + 1) + 3x(x - 1)}{x(x - 1)(x + 1)} = \frac{2x^{2} + 2x + 3x^{2} - 3x}{x^{3} - x^{2} + x}\). This simplifies further to \(\frac{5x^{2} - x}{x^{3} - x^{2} + x}\)

Divide every term by \(x^{3}\)

To find the limit as x approaches negative infinity, divide every term in the rational function by \(x^{3}\) to get: \(\frac{(5x^{2}/x^{3}) - (x/x^{3})}{(x^{3}/x^{3}) - (x^{2}/x^{3}) + (x/x^{3})} = \frac{5/x - 1/x^{2}}{1 - 1/x + 1/x^{2}}\)

Apply the limit

This limit problem can now be solved by applying the concept that the limit as x approaches negative infinity of 1/x is 0. This gives: \(\lim _{x \rightarrow-\infty}\left(\frac{5/x - 1/x^{2}}{1 - 1/x + 1/x^{2}}\right) = \frac{5/(-\infty) - 1/(-\infty)^{2}}{1 - 1/(-\infty) + 1/(-\infty)^{2}} = \frac{0}{1} = 0\)

Key concepts

Limit of a Rational Function

Understanding the limit of a rational function is essential in calculus, especially when analyzing how functions behave as they tend toward infinity. A rational function is a ratio of two polynomials. To find the limit of such a function as the variable, say 'x', approaches a certain value, we need to look at the behavior of the numerator and the denominator separately.

When dealing with limits approaching infinity, one common technique is to divide each term by the highest power of 'x' found in the denominator. This usually simplifies the expression and reveals the function's end behavior. For example, the exercise presented involves a limit as 'x' approaches negative infinity. By dividing each term by the highest power of 'x', which is \(x^3\), the expression can be greatly simplified to show that as 'x' goes to negative infinity, the function will approach a certain value—in this case, zero.

Simplifying Rational Expressions

Rational expressions can often seem complex and difficult to work with. To simplify these expressions and make it easier to find their limits, we combine any like terms and factor wherever possible, essentially reducing the expression to its simplest form. This process involves finding the least common denominator (LCD) when combining fractions.

In the given exercise, the expression is simplified by first combining the two fractions with a common denominator. Once combined, this allows the cancelation of terms and further simplification. Simplifying rational expressions is a fundamental skill in calculus because it facilitates the analysis of functions' behaviors and making the calculation of limits more manageable.

Infinite Limits

In calculus, an infinite limit refers to the value that a function approaches as the variable approaches infinity, either positive or negative. When assessing infinite limits of rational functions, it's often useful to recognize that terms involving lower powers of 'x' become insignificant compared to those with higher powers of 'x' as 'x' grows larger in absolute value.

Applying this concept to the exercise at hand, as 'x' approaches negative infinity, the terms \(\frac{1}{x}\) and \(\frac{1}{x^2}\) approach zero, making them negligible. This often leads to a situation where the function's limit can be determined by examining the leading terms' coefficients or by recognizing that a term becomes insignificant in the limit process. Hence, as 'x' approaches negative infinity in the given function, the result can be determined to be zero.