Over the real line

Calculus Level 3

lim x ( x + ln 9 x ln 9 ) x = ? \large \displaystyle\lim_{x\rightarrow \infty} \left(\dfrac{x + \ln 9}{x - \ln 9}\right)^x = \ ?

0 81 27 18 9

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2 solutions

Marta Reece
Jun 22, 2017

ln ( lim x ( x + ln 9 x ln 9 ) x ) = lim x ln ( x + ln 9 ) ln ( x ln 9 ) 1 x \ln\left(\displaystyle\lim_{x\rightarrow \infty} \left(\dfrac{x + \ln 9}{x - \ln 9}\right)^x\right) =\displaystyle\lim_{x\rightarrow \infty}\dfrac{\ln(x+\ln9)-\ln(x-\ln9)}{\frac1x}

Using L'Hospital's rule, this can be written as

lim x ( x 2 x ln 9 x 2 x + ln 9 ) = lim x 2 ln 9 1 ( ln 9 x ) 2 = ln 81 \displaystyle\lim_{x\rightarrow \infty}\left(\dfrac{x^2}{x-\ln9}-\dfrac{x^2}{x+\ln9}\right)=\displaystyle\lim_{x\rightarrow \infty}\dfrac{2\ln9}{1-\left(\frac{\ln9}x\right)^2}=\ln{81}

Therefore

lim x ( x + ln 9 x ln 9 ) x = 81 \displaystyle\lim_{x\rightarrow \infty} \left(\dfrac{x + \ln 9}{x - \ln 9}\right)^x=\boxed{81}

Zach Abueg
Jun 22, 2017

L = lim x ( x + ln 9 x ln 9 ) x a x = e ln a x = e x ln a = lim x e x ln ( x + ln 9 x ln 9 ) \displaystyle \begin{aligned} \\ \\ L & = \lim_{x \ \to \ \infty} \left(\frac{x + \ln 9}{x - \ln 9}\right)^x & \small \color{#3D99F6} a^x = e^{\ln a^x} = e^{x\ln a} \\ & = \lim_{x \ \to \ \infty} e^{x \ln \left(\frac{x + \ln 9}{x - \ln 9}\right)} \\ \\ \end{aligned}

Now recall the limit chain rule:

If lim x a g ( x ) = b \displaystyle \lim_{x \ \to \ a} g(x) = b and lim u b f ( u ) = L \displaystyle \lim_{u \ \to \ b} f(u) = L , and f ( x ) f(x) is continuous at x = b x = b , then lim x a f ( g ( x ) ) = L \displaystyle \lim_{x \ \to \ a} f\left(g(x)\right) = L .

Let g ( x ) = x ln ( x + ln 9 x ln 9 ) \displaystyle g(x) = x\ln\left(\frac{x + \ln9}{x - \ln9}\right) and f ( u ) = e u \displaystyle f(u) = e^u .

lim x g ( x ) = lim x x ln ( x + ln 9 x ln 9 ) = lim x ln ( x + ln 9 x ln 9 ) 1 x Apply L’hopital’s Rule = lim x 4 ln 3 ( x + 2 ln 3 ) ( x 2 ln 3 ) 1 x 2 = lim x 4 x 2 ln 3 x 2 4 ln 2 3 Apply L’hopital again = lim x 8 x ln 3 2 x = lim x 4 ln 3 = 4 ln 3 lim u 4 ln 3 f ( u ) = lim x 4 ln 3 e u = 81 \displaystyle \begin{aligned} \\ \lim_{x \ \to \ \infty} g(x) & = \lim_{x \ \to \ \infty} x\ln\left(\frac{x + \ln9}{x - \ln9}\right) \\ & = \lim_{x \ \to \ \infty} \frac{\ln\left(\frac{x + \ln9}{x - \ln9}\right)}{\frac 1x} & \small \color{#3D99F6} \text{Apply L'hopital's Rule} \\ & = \lim_{x \ \to \ \infty} \frac{- \frac{4\ln3}{\left(x + 2\ln3\right)\left(x - 2\ln3\right)}}{-\frac{1}{x^2}} \\ & = \lim_{x \ \to \ \infty} \frac{4x^2\ln3}{x^2 - 4\ln^23} & \small \color{#3D99F6} \text{Apply L'hopital again} \\ & = \lim_{x \ \to \ \infty} \frac{8x\ln3}{2x} \\ & = \lim_{x \ \to \ \infty} 4\ln3 \\ & = 4\ln3 \\ \\ \lim_{u \ \to \ 4\ln3} f(u) & = \lim_{x \ \to \ 4\ln3} e^u \\ & = \boxed{81} \end{aligned}

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