Question

If $\lim _{x \rightarrow \infty} \frac{729^{x}-243^{x}-81^{x}+9^{x}+3^{x}-1}{x^{3}}=k(\ln 3)^{3}\(, then \)k$ is equal to (A) 4 (B) 5 (C) 6 (D) none

Short answer

Question: Find the value of k for which the limit is true: \(\lim _{x \rightarrow \infty}\frac{(3^6)^{x}-(3^5)^{x}-(3^4)^{x}+(3^2)^{x}+(3^1)^{x}-1}{x^{3}} = k(\ln 3)^3\) Answer: \(k = 6\)

Step-by-step solution

Write the expression in a neater form

To make the expression more workable, let's rewrite it using properties of exponents: \(\lim _{x \rightarrow \infty}\frac{(3^6)^{x}-(3^5)^{x}-(3^4)^{x}+(3^2)^{x}+(3^1)^{x}-1}{x^{3}}\)

Factor out a common power of 3 from the numerator

Notice that \((3^1)^{x}\) is the smallest power of 3 in the numerator, so let's factor out \(3^{x}\) from all the terms: \(\lim _{x \rightarrow \infty}\frac{3^x(3^{5x}-3^{4x}-3^{3x}+3^x+1-3^{-x})}{x^3}\)

Change the expression inside the limit into a sum of fractions and make a substitution

Divide both the numerator and denominator by \(3^x\): \(\lim _{x \rightarrow \infty}\frac{3^{5x}-3^{4x}-3^{3x}+3^x+1-3^{-x}}{x^3}\) Let's set \(y = 3^x\), then as \(x \rightarrow \infty\), \(y \rightarrow \infty\). Now let's rewrite the limit: \(\lim _{y \rightarrow \infty}\frac{y^5-y^4-y^3+y+1-\frac{1}{y}}{(\ln 3)^3(\ln y)^3}\)

Divide both the numerator and denominator by the highest power of y

Dividing both the numerator and denominator by \(y^5\) we get: \(\lim _{y \rightarrow \infty}\frac{1-\frac{1}{y}-(\frac{1}{y})^2+(\frac{1}{y})^4+(\frac{1}{y})^5-\frac{1}{y^{6}}}{(\ln 3)^3 (\frac{\ln y}{y})^3}\)

Apply L'Hôpital's Rule

Since we have an indeterminate form \(\frac{0}{0}\) as \(y \rightarrow \infty\), we can apply L'Hôpital's Rule 3 times to the transformed expression: \(\lim _{y \rightarrow \infty}\frac{6}{(\ln 3)^3 (\frac{-2}{y^2})^3} = k (\ln 3) ^3\) As a result, we have: \(k (\ln 3) ^3 = \frac{6}{(\ln 3)^3}\) \(k = \boxed{6\ (C)}\)