1. ## L'Hospital

$\displaystyle \lim_{x \to \infty} (\frac{x-4}{x+5})^(x-2)$

2. Originally Posted by Apprentice123
$\displaystyle \lim_{x \to \infty} (\frac{x-4}{x+5})^(x-2)$
Why not show some work

$\displaystyle \ln(y)=\lim_{x\to\infty}(x-2)\ln\left(\frac{x-4}{x+5}\right)$

3. $\displaystyle ln \lim_{x \to \infty} y=\lim_{x \to \infty} (x-2).\frac{(x+5)}{(x-4)}.\frac{1}{(x+5)^2}$

$\displaystyle ln \lim_{x \to \infty} y=\lim_{x \to \infty} \frac{x^2+5x-2x-10}{x^3+10x^2+25x-4x^2-40x-100}$

$\displaystyle ln \lim_{x \to \infty} y=\lim_{x \to \infty} \frac{x^2+3x-10}{x^3+6x^2+15x-100}$

$\displaystyle ln \lim_{x \to \infty} y=\lim_{x \to \infty} \frac{2}{6x}$

Its new?

The answer is:$\displaystyle \frac{1}{e^9}$

4. Originally Posted by Apprentice123
$\displaystyle \lim_{x \to \infty} (\frac{x-4}{x+5})^(x-2)$
Let $\displaystyle L=\lim_{x \to \infty} (\frac{x-4}{x+5})^{x-2}$

Taking the natural log of both sides, we have

\displaystyle \begin{aligned} \ln(L)&=\lim_{x\to{\infty}}(x-2)\ln\bigg(\frac{x-4}{x+5}\bigg) \\ &=\lim_{x\to{\infty}}\frac{\ln\bigg(\frac{x-4}{x+5}\bigg)}{\frac{1}{x-2}}\end{aligned}

Applying L'Hopital's Rule, we get

$\displaystyle \lim_{x\to{\infty}}\frac{\frac{x+5}{x-4}\cdot\frac{9}{(x+5)^2}}{-\frac{1}{(x-2)^2}}$

Rewriting, we get:

$\displaystyle \lim_{x\to{\infty}}\frac{-9(x-2)^2}{(x-4)(x+5)}=\color{blue}\boxed{-9}$

But recall that $\displaystyle \ln(L)=\lim_{x\to{\infty}}(x-2)\ln\bigg(\frac{x-4}{x+5}\bigg)$

Thus, $\displaystyle \ln(L)=-9\implies \color{blue}\boxed{L=e^{-9}}$

I hope that this makes sense!

--Chris

5. yes. Thank you