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In Exercises \(37-42,\) find the limit of each function (a) as \(x \rightarrow \infty\) and (b) as \(x \rightarrow-\infty\) . (You may wish to visualize your answer with a graphing calculator or computer.) $$ h(x)=\frac{-5+(7 / x)}{3-\left(1 / x^{2}\right)} $$

Short Answer

Expert verified
The limit is \(-\frac{5}{3}\) as \(x\rightarrow\infty\) and \(x\rightarrow-\infty\).

Step by step solution

01

Simplify for Limits as x approaches Infinity

To find the limit as \( x \rightarrow \infty \), first simplify the terms with \( x \) in the denominator. The expression for \( h(x) \) can be rewritten by factoring out the highest power of \( x \) from both the numerator and denominator:\[h(x) = \frac{-5 + \frac{7}{x}}{3 - \frac{1}{x^2}}\]As \( x \) approaches infinity, terms \( \frac{7}{x} \) and \( \frac{1}{x^2} \) approach 0. Therefore, the function simplifies to:\[\lim_{x \to \infty} h(x) = \frac{-5 + 0}{3 - 0} = \frac{-5}{3}\]
02

Simplify for Limits as x approaches Negative Infinity

Next, find the limit as \( x \rightarrow -\infty \). Simplify the expression in the same way as in Step 1:\[ h(x) = \frac{-5 + \frac{7}{x}}{3 - \frac{1}{x^2}} \]Again, as \( x \) approaches negative infinity, \( \frac{7}{x} \) and \( \frac{1}{x^2} \) tend toward 0. The simplified expression is then:\[\lim_{x \to -\infty} h(x) = \frac{-5 + 0}{3 - 0} = \frac{-5}{3}\]
03

Conclusion: State the Limits at Infinity and Negative Infinity

For both cases as \( x \to \infty \) and \( x \to -\infty \), the limits of the function \( h(x) \) converge to the same value. Therefore, the limit in both cases is:\[\frac{-5}{3}\]This indicates the horizontal asymptote of the function at a large positive or negative \( x \).

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Key Concepts

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

Horizontal Asymptote
A horizontal asymptote is a horizontal line that a function approaches as the input, usually denoted by \( x \), becomes very large or very small. In simpler terms, as you move far to the right or far to the left along the x-axis on a graph, the value of the function gets closer and closer to the horizontal asymptote.
In the context of the function \( h(x) = \frac{-5 + \frac{7}{x}}{3 - \frac{1}{x^2}} \), we've calculated that as \( x \to \infty \) and \( x \to -\infty \), \( h(x) \) approaches \( \frac{-5}{3} \).
This means the horizontal asymptote of the function is the line \( y = \frac{-5}{3} \).
  • The points \( \frac{7}{x} \) and \( \frac{1}{x^2} \) weaken as \( x \) increases or decreases extensively because they are divided by large numbers.
  • The main terms \( -5 \) and \( 3 \) remained constant, leading the value towards \( \frac{-5}{3} \).
This helps to tell us how the function behaves for values that are way out on either side of the graph.
Infinity
Infinity is not a number, but a concept. It's like saying something is immeasurably large. Understanding infinity in mathematics helps us determine what happens to a value when it grows without bounds or shrinks beyond measurable limits.
When we say \( x \to \infty \), it means that \( x \) increases towards larger and larger positive values.
For \( x \to -\infty \), \( x \) is instead moving towards larger negative values.
In our exercise, when calculating the limit of \( h(x) \) as \( x \to \infty \) or \( x \to -\infty \), we simplify the formula by neglecting terms with \( x \) in the denominator. This is because their impact diminishes to zero as \( x \) approaches infinity.
  • As \( \frac{7}{x} \to 0 \), the fraction diminishes since 7 is divided by an enormous number.
  • Similarly, \( \frac{1}{x^2} \to 0 \) occurs even faster as any large number squared becomes even larger.
Thus, understanding infinity helps identify how certain terms in a function cease to contribute significantly as \( x \) moves towards extremely large or small values.
Rational Functions
Rational functions are expressions that involve fractions, with both the numerator and the denominator being polynomials.
The function \( h(x) = \frac{-5 + \frac{7}{x}}{3 - \frac{1}{x^2}} \) is an example of a rational function. Here, we're dealing with terms \( -5 + \frac{7}{x} \) in the numerator and \( 3 - \frac{1}{x^2} \) in the denominator.
In rational functions, a key feature is their behavior at values of \( x \) where the denominator may become zero, unless specified otherwise, would result in undefined values.
Understanding how to find limits for rational functions involves:
  • Simplifying by ignoring parts of the formula that become negligible at extreme values of \( x \).
  • Recognizing dominant terms that dictate the function's behavior at infinity or negative infinity.
In our case, all polynomial terms divided by \( x \) or \( x^2 \) diminish, letting the prevailing constants dictate the approach to the horizontal asymptote. This simplifies calculation and helps readily understand how rational functions change over very large or very small inputs.

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Most popular questions from this chapter

Graph the rational functions in Exercises \(27-38 .\) Include the graphs and equations of the asymptotes and dominant terms. $$ y=\frac{x^{2}+1}{x-1} $$

Does the graph of $$ g(x)=\left\\{\begin{array}{ll}{x \sin (1 / x),} & {x \neq 0} \\ {0,} & {x=0}\end{array}\right. $$ have a tangent at the origin? Give reasons for your answer.

In Exercises \(61-66,\) you will further explore finding deltas graphically. Use a CAS to perform the following steps: a. Plot the function \(y=f(x)\) near the point \(x_{0}\) being approached. b. Guess the value of the limit \(L\) and then evaluate the limit symbolically to see if you guessed correctly. c. Using the value \(\epsilon=0.2,\) graph the banding lines \(y_{1}=L-\epsilon\) and \(y_{2}=L+\epsilon\) together with the function \(f\) near \(x_{0}\) . d. From your graph in part (c), estimate a \(\delta > 0\) such that for all \(x\) $$ 0 < \left|x-x_{0}\right| < \delta \quad \Rightarrow \quad|f(x)-L| < \epsilon $$ Test your estimate by plotting \(f, y_{1},\) and \(y_{2}\) over the interval \(0 < \left|x-x_{0}\right| < \delta .\) For your viewing window use \(x_{0}-2 \delta \leq x \leq x_{0}+2 \delta\) and \(L-2 \epsilon \leq y \leq L+2 \epsilon\) . If any function values lie outside the interval \([L-\epsilon, L+\epsilon],\) your choice of \(\delta\) was too large. Try again with a smaller estimate. e. Repeat parts (c) and (d) successively for \(\epsilon=0.1,0.05,\) and \(0.001 .\) $$ f(x)=\frac{3 x^{2}-(7 x+1) \sqrt{x}+5}{x-1}, \quad x_{0}=1 $$

Each of the functions in Exercises \(57-60\) is given as the sum or difference of two terms. First graph the terms (with the same set of axes). Then, using these graphs as guides, sketch in the graph of the function. $$ y=\frac{1}{x}-\tan x, \quad-\frac{\pi}{2} < x < \frac{\pi}{2} $$

Suppose that \(f\) is an even function of \(x\) . Does knowing that \(\lim _{x \rightarrow 2^{-}} f(x)=7\) tell you anything about either \(\lim _{x \rightarrow-2^{-}} f(x)\) or \(\lim _{x \rightarrow-2^{+}} f(x) ?\) Give reasons for your answer.

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