How do I apply L'Hospital's rule here?

I need to compute the limit of $\left(\frac{10x}{10x+3}\right) ^{6x}$ as $x$ approaches infinity. I know I need to use L'Hospital's rule and I feel like I need to use logarithms first before I apply L'Hospital's rule but i'm not sure how to go about doing that.

Yes, you compute the limit of the logarithm of the expression: $$\lim_{x\to\infty}6x\log\dfrac{10x}{10x+3}$$ Then the limit of the original function is the exponential of this limit.

The easiest way is to do $x=1/t$, so you get $$\lim_{t\to0^+}\frac{6}{t}\log\frac{10}{10+3t}= 6\lim_{t\to0^+}\frac{\log10-\log(10+3t)}{t}$$ Can you go on?

A different strategy is to note that $$\frac{10x}{10x+3}=\frac{10x+3-3}{10x+3}=1-\frac{3}{10x+3}$$ Now set $10x+3=t$, so $6x=\frac{3}{5}(t-3)$ and the limit is $$\lim_{t\to\infty}\left(\left(1-\frac{3}{t}\right)^{t}\left(1-\frac{3}{t}\right)^{-3}\right)^{3/5}=(e^{-3}\cdot 1)^{3/5}$$

• isn't the limit of the expression $\lim_{x\to\infty}6x\log\dfrac{10x}{10x+3}$ the limit of $lny$ as $x$ approaches infinity and not the limit of $y$ as $x$ approaches infinity? Or are they both the same? Apr 1 '16 at 22:01
• @user140161 No they're not the same: you get $-9/5$ as the limit of $\log y$, so the limit of $y$ is $e^{-9/5}$. Apr 1 '16 at 22:03

Making $h=\frac{3}{10x}$ you have $$\left(\frac{10x}{10x+3}\right)^{6x}=\left(1+\frac{3}{10x}\right)^{-6x}=(1+h)^{-6x}$$ Hence $$\lim_{x\to 0}\left((1+h)^{\frac 1h}\right)^{-6hx}=\lim_{x\to 0}e^{-\frac{18x}{10x}}=\color{red}{e^{-\frac 95}}$$

• @140161: What I wanted to say you is you do not need necessarily L'Hospital's rule. Apr 1 '16 at 22:39