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Chapter 1: Problem 71

In Problems \(64-71\), find a value of the constant \(k\) such that the limit exists. $$\lim _{x \rightarrow-\infty} \frac{3^{k x}+6}{3^{2 x}+4}$$

Short Answer

Expert verified
For the limit to exist, \(k = 2\).

Step by step solution

01

Analyze the Dominant Terms

When dealing with limits at infinity, identify which terms grow fastest as \(x\) approaches \(-\infty\). In the numerator, the dominant term is \(3^{kx}\), and in the denominator, it is \(3^{2x}\). These terms will have the greatest influence on the behavior of the expression as \(x\) approaches \(-\infty\).
02

Simplify the Expression by Factoring Out Dominant Terms

We simplify by factoring out the dominant terms. This means rewriting the expression: \(\lim _{x \rightarrow-\infty} \frac{3^{kx}(1 + \frac{6}{3^{kx}})}{3^{2x}(1 + \frac{4}{3^{2x}})}\). Notice that the \(\frac{6}{3^{kx}}\) and \(\frac{4}{3^{2x}}\) terms will approach zero as \(x\) goes to \(-\infty\).
03

Simplify Further with Limits Approaching Zero

Under the limit, the fractions \(\frac{6}{3^{kx}}\) and \(\frac{4}{3^{2x}}\) become negligible as they tend to zero. Therefore, the expression simplifies to: \(\lim _{x \rightarrow-\infty} \frac{3^{kx}}{3^{2x}}\).
04

Determine the Simplified Limit Expression

The simplified limit expression is now \(3^{kx-2x} = 3^{(k-2)x}\). For the limit to exist as \(x\) approaches \(-\infty\), the exponent must be zero, because a very small value (negative infinity) should not lead the function to grow to zero or infinity.
05

Solve for \(k\) to Make the Limit Exist

Set the exponent \((k-2)\) equal to zero: \(k - 2 = 0\), solving gives \(k = 2\). Thus, for the limit to exist, \(k\) must be 2.

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Key Concepts

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

Dominant Terms
When calculating limits, especially as\( x \) approaches infinity or negative infinity, it's crucial to identify the dominant terms in the expression. Dominant terms are those that grow the fastest and have the most significant impact on the behavior of a function. Consider the function
  • Numerator: \( 3^{kx} + 6 \)
  • Denominator: \( 3^{2x} + 4 \)
As \( x \) goes to \(-\infty\), the term \( 3^{kx} \) dominates over the constant 6 in the numerator, and \( 3^{2x} \) dominates over the constant 4 in the denominator.
The key is to focus on these dominant terms for simplification and further calculations.
Simplification of Expressions
To simplify expressions involving limits, especially at infinities, we often factor out the dominant terms. This process makes it easier to address and simplify the expression step by step.
In our particular example:
  • We have \( \lim_{x \rightarrow -\infty} \frac{3^{kx}(1 + \frac{6}{3^{kx}})}{3^{2x}(1 + \frac{4}{3^{2x}})} \).
The terms \( \frac{6}{3^{kx}} \) and \( \frac{4}{3^{2x}} \) approach zero as \( x \) approaches \(-\infty\), effectively simplifying the expression to \( \frac{3^{kx}}{3^{2x}} \).
This simplification helps make the complex expression more manageable and paves the way for applying limit calculations.
Limit at Infinity
Limits at infinity articulate how a function behaves as \( x \) approaches infinitely large or infinitely small values. In this context, understanding the growth rates of exponential terms is key.
After simplification,
  • The expression \( \lim_{x \rightarrow -\infty} \frac{3^{kx}}{3^{2x}} \) becomes apparent.
This can be rewritten as \( 3^{(k-2)x} \), providing a clear structure to evaluate the behavior of the function.
For the limit to exist, the exponential term's growth or decay should not lead to an infinite value鈥攈ence the need for further exploration.
Exponent Equalization
Exponent equalization is an essential step in ensuring the calculated limit exists. With exponential functions like \( 3^{(k-2)x} \), the growth order depends on the exponent.
For the given problem, setting the exponent \( k-2 \) equals zero is vital. This step ensures the expression stabilizes:
  • The equation becomes \( k - 2 = 0 \)
  • Solving this gives \( k = 2 \).
Finding \( k = 2 \) means the function will not head towards infinity or zero as \( x \rightarrow -\infty \), allowing the existence of the limit.
This outcome ensures that the limiting behavior of the function is steady and comprehensible.

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Most popular questions from this chapter

Give an example of: A formula representing the statement " \(q\) is inversely proportional to the cube root of \(p\) and has a positive constant of proportionality."

Give an example of: A cosine function which oscillates between values of 1200 and 2000

Consider the function \(f(x)=\sin (1 / x)\) (a) Find a sequence of \(x\) -values that approach 0 such that \(\sin (1 / x)=0\) [Hint: Use the fact that \(\sin (\pi)=\sin (2 \pi)=\) \(\sin (3 \pi)=\ldots=\sin (n \pi)=0 .]\) (b) Find a sequence of \(x\) -values that approach 0 such that \(\sin (1 / x)=1\) IHint: Use the fact that \(\sin (n \pi / 2)=1\) if \(n=\) \(1,5,9, \dots .]\) (c) Find a sequence of \(x\) -values that approach 0 such that \(\sin (1 / x)=-1\) (d) Explain why your answers to any two of parts (a)-(c) show that \(\lim _{x \rightarrow 0} \sin (1 / x)\) does not exist.

For the functions in Problems \(76-77\), do the following: (a) Make a table of values of \(f(x)\) for \(x=a+0.1, a+0.01\) \(a+0.001, a+0.0001, a-0.1, a-0.01, a-0.001\) and \(a-0.0001\) (b) Make a conjecture about the value of \(\lim _{T \rightarrow a} f(x)\) (c) Graph the function to see if it is consistent with your answers to parts (a) and (b). (d) Find an interval for \(x\) containing \(a\) such that the difference between your conjectured limit and the value of the function is less than 0.01 on that interval. (In other words, find a window of height 0.02 such that the graph exits the sides of the window and not the top or bottom of the window.) $$f(x)=\frac{\cos 2 x-1+2 x^{2}}{x^{3}}, \quad a=0$$

When Galileo was formulating the laws of motion, he considered the motion of a body starting from rest and falling under gravity. He originally thought that the velocity of such a falling body was proportional to the distance it had fallen. What do the experimental data in Table 1.6 tell you about Galileo's hypothesis? What alternative hypothesis is suggested by the two sets of data in Table 1.6 and Table \(1.7 ?\) $$\begin{array}{c|c|c|c|c|c}\hline \text { Distance (ft) } & 0 & 1 & 2 & 3 & 4 \\\\\hline \text { Velocity (ft/sec) } & 0 & 8 & 11.3 & 13.9 & 16 \\\\\hline\end{array}$$ $$\begin{array}{c|c|c|c|c|c}\hline \text { Time (sec) } & 0 & 1 & 2 & 3 & 4 \\\\\hline \text { Velocity (ft)sec) } & 0 & 32 & 64 & 96 & 128 \\\\\hline\end{array}$$
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