L'hopital's Rule on an indeterminate difference I have the following question, in which I am told to use L'Hopital's rule:
$\lim_{x\to \infty}(-x+5\cdot ln(x))$
From eyeballing it, I would conclude that the polynomial $x$ will decrease faster than the logarithmic $ln$ would increase, meaning the limit would be $-\infty$, but I can't see how to use L'hopitals rule here as I've been told. I know that the idea is to put this into a quotient, but if I do the following:
$$-x+5\cdot ln(x) = \frac{-x^2}{x}+\frac{5x\cdot ln(x)}{x} = \frac{-x^2 + 5x\cdot ln(x)}{x}$$
And then try to take the limit, I get a doubly indeterminate form of $\frac{-\infty+\infty}{\infty}$, and using L'Hopitals rule here just ends back inthe same kind of indeterminate difference.
 A: Try exponentiating; write $$-x+5\ln x=5\ln\left(e^{-x/5+\ln x}\right)=5\ln\left(\frac{x}{e^{x/5}}\right),$$ and apply L'Hopital's rule at the argument of the logarithm.
A: $$\begin{array}{lll}
\displaystyle\lim_{x\to \infty}(-x+5\ln x)&=&\displaystyle\lim_{x\to \infty}(-x+\ln x^5)\\
&=&\displaystyle\lim_{x\to \infty}(\ln(e^{-x+\ln x^5}))\\
&=&\ln\displaystyle\lim_{x\to \infty}(e^{-x+\ln x^5})\\
&=&\ln\displaystyle\lim_{x\to \infty}(e^{-x}e^{\ln x^5})\\
&=&\ln\displaystyle\lim_{x\to \infty}(x^5e^{-x})\\
&=&\ln\displaystyle\lim_{x\to \infty}\bigg(\frac{x^5}{e^x}\bigg)\\
&=&\dots\\
\end{array}$$
A: Hint: you might want to start with $-x+5\ln(x)=x\cdot\left(-1+\frac{5\ln(x)}{x}\right)$. Clearly $\lim\limits_{x\to\infty} x= \infty$, so if you can show that $\lim\limits_{x\to\infty} \left(-1+\frac{5\ln(x)}{x}\right)\neq 0$ you should have no trouble concluding that $\lim\limits_{x\to\infty} \left(-x+5\ln(x)\right)=-\infty$.
Edit to clarify: from $\lim\limits_{x\to\infty} \left(-1+\frac{5\ln(x)}{x}\right)\neq 0$ one doesn't immediately get $\lim\limits_{x\to\infty} \left(-x+5\ln(x)\right)=-\infty$, but as $\lim\limits_{x\to\infty} \left(-1+\frac{5\ln(x)}{x}\right)\in\mathbb R$ exists, this will lead to the solution.
