Problem 71 - Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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 \(-\infty\).

Step 2: 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 \(-\infty\).

Step 3: 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: \(\lim _{x \rightarrow-\infty} \frac{3^{kx}}{3^{2x}}\).

Step 4: 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.

Step 5: 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.

Key Concepts

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—hence 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.