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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.

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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 $$ \lim_{t\to\infty}\left(\left(1-\frac{3}{t}\right)^{t}\left(1-\frac{3}{t}\right)^{-3}\right)^{3/5}=(e^{-3}\cdot 1)^{3/5} $$

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  • $\begingroup$ 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? $\endgroup$ – user140161 Apr 1 '16 at 22:01
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    $\begingroup$ @user140161 No they're not the same: you get $-9/5$ as the limit of $\log y$, so the limit of $y$ is $e^{-9/5}$. $\endgroup$ – egreg Apr 1 '16 at 22:03
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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 $$\lim_{x\to 0}\left((1+h)^{\frac 1h}\right)^{-6hx}=\lim_{x\to 0}e^{-\frac{18x}{10x}}=\color{red}{e^{-\frac 95}}$$

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

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