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I am having a little trouble going about: $$\lim_{x\to \infty} \left(\frac{14x}{14x+10}\right)^{10x}$$

Using $\ln$ properties we can bring down the $10x$ exponent and have:$$\ln y=10x\ln\left(\frac{14x}{14x+10}\right)$$

And from here I get stuck trying to apply L'Hospital's Rule to find the limit.

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  • $\begingroup$ You don't necessarly need l'Hopital here and I suggest to avoid that when we can proceed by standard limits. $\endgroup$ – gimusi Dec 4 '18 at 20:35
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Hint :

$$10x \ln \left(\frac{14x}{14x+10}\right)= \frac{\ln \left(\frac{14x}{14x+10}\right)}{\frac{1}{10x}}$$

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HINT

We can use that

$$\left(\frac{14x}{14x+10}\right)^{10x}=\left(\frac{14x+10-10}{14x+10}\right)^{10x}=\left[\left(1-\frac{10}{14x+10}\right)^{\frac{14x+10}{10}}\right]^{\frac{10\cdot 10x}{14x+10}}$$

and refer to standard limits.

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  • $\begingroup$ +1 by me since that was the original answer manipulating the definition of $e$. $\endgroup$ – Rebellos Dec 4 '18 at 21:42
  • $\begingroup$ @Rebellos Thanks also your hint is very effective since the asker was looking for a method to apply l'Hopital rule! :) $\endgroup$ – gimusi Dec 4 '18 at 21:45
  • $\begingroup$ I still don't get where e can come from here??? $\endgroup$ – pijoborde Dec 4 '18 at 21:56
  • $\begingroup$ @pijoborde Are you aware for as $n\to \infty$ $\left(1+\frac1n\right)^n \to e$ and $\left(1+\frac a n\right)^n \to e^a$? $\endgroup$ – gimusi Dec 4 '18 at 21:58
  • $\begingroup$ And our a value is represented by 10 in this instance or? Making the answer e raised to 10? I mean, I feel like the answer is supposed to be e raised to some negative fraction. $\endgroup$ – pijoborde Dec 4 '18 at 22:02
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Hint: You want to reach some kind of indeterminate form, so you need to mess around with the fractions to reach a desirable form.

$$\ln y=10x\ln\left(\frac{14x}{14x+10}\right)$$

$$\ln y=\frac{\ln\left(\frac{14x}{14x+10}\right)}{\frac{1}{10x}}$$

Now, you can continue.


As an alternative approach (both faster and easier), you can use $\lim_\limits{n \to \infty}\big(1+\frac{x}{n}\big)^n = e^x$. These expressions are easily manipulated to reach such a form.

Notice the numerator is $10$ less than the denominator, so $\frac{14x}{14x+14} = 1+\frac{-10}{14x+10}$. Hence, you get

$$\lim_{x\to \infty} \left(\frac{14x}{14x+10}\right)^{10x} = \lim_{x\to \infty} \Biggl[\left(1+\frac{-10}{14x+10}\right)^{14x+10}\Biggl]^{\frac{10x}{14x+10}}$$

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  • $\begingroup$ (+1) for the double hint $\endgroup$ – gimusi Dec 4 '18 at 20:34
  • $\begingroup$ I am supposed to get an answer involving e. So I need to focus on your alternative approach method? $\endgroup$ – pijoborde Dec 4 '18 at 20:45
  • $\begingroup$ @pijoborde You can use both method. I prefer and suggest the second one which is much more instructive. $\endgroup$ – gimusi Dec 4 '18 at 20:46
  • $\begingroup$ I don't know if you've worked on the limit provided for $e^x$. If you're allowed to use it, why not? For all limits like this, just look for a way to reach something in the form $$\lim_{n \to \infty}\bigg[\bigg(1+\frac{x}{n}\bigg)^n\bigg]^m$$ and evaluate the limit from there. It's very simple and has little room for error. $\endgroup$ – KM101 Dec 4 '18 at 20:47
  • $\begingroup$ Your answers are complete, but in my honest opinion, I don't find much sense in answering the same thing as others have first done and just elaborating it. The first part of your answer was answered by me and the second part by Gimusi. $\endgroup$ – Rebellos Dec 4 '18 at 21:42
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Once you get to the logarithm, you need $$ \lim_{x\to\infty}\frac{\ln\dfrac{14x}{14x+10}}{\dfrac{1}{10x}} $$ which is in the form $0/0$ and l'Hôpital can be applied. However it's much simpler if the numerator is written as $\ln14+\ln x-\ln(14x+10)$, in order to compute the derivative (you know the limit is $0$ anyhow). So we get $$ \lim_{x\to\infty}\frac{\dfrac{1}{x}-\dfrac{14}{14x+10}}{-\dfrac{1}{10x^2}}= \lim_{x\to\infty}-10x^2\frac{14x+10-14x}{x(14x+10)}=\lim_{x\to\infty}-\frac{100x}{14x+10} $$ When you have this limit, let me call it $l$, the one you started with is $e^l$.

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  • $\begingroup$ What happened exactly in the last step? I don't quite get where 100 came from? Unless you are reducing an x from both top and bottom to 1 and then distributing the -10x to the 10, since the positive and minus 14x will cancel out? $\endgroup$ – pijoborde Dec 6 '18 at 7:01
  • $\begingroup$ @pijoborde $10\cdot10=100$ $\endgroup$ – egreg Dec 6 '18 at 9:01
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Tips:

With equivalents, it's very fast:$$\ln\biggl(\frac{14x}{14x+10}\biggr)=\ln\biggl(1-\frac{10}{14x+10}\biggr)\sim_\infty -\frac{10}{14x+10}\sim_\infty -\frac{10}{14x}=-\frac 57,$$ so, as equivalence is compatible with multiplication/division, $$10x\ln\biggl(\frac{14x}{14x+10}\biggr)\sim_\infty -\frac{50x}{7x}\to -\frac{50}7.$$ Warning: equivalence is not compatible with addition/subtraction. It is compatible, under mild hypotheses, with composition on the left by logarithm.

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    $\begingroup$ Equivalents are fast but also dangerous to handle for limits. $\endgroup$ – gimusi Dec 4 '18 at 20:33
  • $\begingroup$ Why? If one knows the rules… When I was a student, our professor used to say that L'Hospital's rule can be dangerous (there are some hypotheses no one ever thinks to check), and when it works, Taylor at order $1$ works as well. $\endgroup$ – Bernard Dec 4 '18 at 20:39
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    $\begingroup$ I'm not referring of course to the use by an expert user like you but for not expert user it can be dangerous. Let consider for example the limit $\frac{t-\log(1+t)}{t^2}$ accorging to $\log (1+t) \sim t$ one could conclude that $$\frac{t-\log(1+t)}{t^2}\sim \frac{t-t}{t^2}=0$$ which is of course wrong. $\endgroup$ – gimusi Dec 4 '18 at 20:44
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    $\begingroup$ I think you should always indicates that issue when you suggest to use equivalents to not expert users. $\endgroup$ – gimusi Dec 4 '18 at 20:45
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    $\begingroup$ I'm not claiming that the method is wrong, I'm claiming that we need to be aware about it to handle them and a not expert user can easily make a mistake. I would suggest for example use them in the form $\log(1+t)=t+o(t)$ in order to take into account the rermainder term. But it is only a suggestion of course. $\endgroup$ – gimusi Dec 4 '18 at 20:55

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