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Is the following method correct? I know you cannot replace an expression with a sub-expression in a limit, but is this the case? I am talking specifically about lines $2$ and $3$. If this is not the correct way, why is my answer correct? How should I approach such limit? Thank you for any help!$$\lim_{u\to \infty}\left(-\frac{u^2\ln\left(u^2\right)}{1+u^2}-\ln\left(\frac1{1+u^2}\right)\right)$$ $$=\lim_{u\to \infty}\left(-\frac{u^2\ln\left(u^2\right)}{u^2\left(1+\frac1{u^2}\right)}+\ln\left(1+u^2\right)\right)$$ $$=\lim_{u\to \infty}\left(-\ln\left(u^2\right)+\ln\left(1+u^2\right)\right)$$ $$=\lim_{u\to \infty}\ln\left(\frac{1+u^2}{u^2}\right)$$ $$=\lim_{u\to \infty}\ln\left(\frac{u^2\left(1+\frac1{u^2}\right)}{u^2}\right)$$ $$=\lim_{u\to \infty}\ln1=0$$ $$$$***Edit: Is there a way to evaluate this limit $\textbf{without}$ the use of Series Expansion? $$$$Maybe this might be useful, the original limit was: $$\lim_{x\to 0}\left(\frac{\ln\left(x^2\right)}{1+x^2}-\ln\left(\frac{x^2}{1+x^2}\right)\right)$$ As you can see, I let: $$x=\frac1{u}$$ As I thought it would be nicer to work with $u\to\infty$

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    $\begingroup$ The proof is wrong since, as you noted, step $3$ does not follow from $2\,$. If this is not the correct way, why is my answer correct? Because after the wrong step anything can happen, including getting the right result by accident, yet the end result alone doesn't make a proof valid. $\endgroup$ – dxiv Mar 3 '17 at 4:29
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To address your initial question: no, your approach isn't correct, for precisely the reason you stated yourself. It just so happens that this wrong approach gives you the right answer.

Anyway, on towards tackling the limit: $$\frac{\ln x^2}{1+x^2} - \ln x^2 + \ln (1+x^2) = \ln x^2 \left(\frac{1}{1+x^2} - 1\right) + \ln (1+x^2)$$

But $\ln(1+x^2) \to \ln 1 = 0$ as $x\to 0$ by continuity of $\log$. So we need only work on $$\frac{x^2\ln x^2}{1+x^2} = \frac{\ln x^2}{1 + x^{-2}}$$

We could potentially use L'Hopital (since it's of the form $\infty/\infty$) on this to get $$\frac{2x^{-1}}{-2x^{-3}} =-x^2 \to 0$$ so $(x^2\ln x^2)/(1+x^2) \to 0$ so the entire expression goes to $0$, since it's the sum of two things that go to $0$. I'm not sure how to avoid L'Hopital at that last step, but I figured I'd post this anyway, especially re: your last comment.

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  • $\begingroup$ It looks good! Do you think there is a way without Hospital? I am gonna wait to see for other answers before accepting yours. However, thank you for the response. $\endgroup$ – DMH16 Mar 3 '17 at 5:22
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    $\begingroup$ Ah, yes: as per the other answer $x^2 \ln x \to 0$ and $1+x^2 \to 1$ so the fraction $x^2 \ln x / (1+x^2) \to 0/1 = 0$. $\endgroup$ – Zain Patel Mar 3 '17 at 5:25
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    $\begingroup$ Note: limits don't go to $0$ they may equal $0$ however $\endgroup$ – zhw. Mar 3 '17 at 5:48
  • $\begingroup$ I just noticed and fixed an error in your derivation where you drop the square from "$\ln(x^2)$" That is not correct because we only have $x \to 0$ and not $x \to 0^+$. $\endgroup$ – user21820 Mar 3 '17 at 17:19
  • $\begingroup$ Please see my post for more detail on the error that I've just fixed in yours. $\endgroup$ – user21820 Mar 3 '17 at 17:32
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As dxiv pointed it out, step 3 does not follow from step 2 (basically you took the limit in the denominator, without making $u$ tends to $-\infty$ in the numerator, which you aren't allowed to).

One way to find your limit without series expansion (so sad) and without L'Hospital rule would be the following:

$$\dfrac{\ln(x^2)}{1+x^2}-\ln\left(\dfrac{x^2}{1+x^2}\right) = -\dfrac{\ln(x^2) x^2}{1+x^2} + \ln(1+x^2)$$

Then $\ln(x^2)x^2 \rightarrow 0$ when $x$ goes to $0$, that's high school stuff (let $y=x^2$, you have to find the limit of $y \ln(y)$ when $y$ goes to $0$). And $1+x^2 \rightarrow 1$. Then your expression goes to $-0/1 + \ln(1)$, which is $0$.

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  • $\begingroup$ Well, to evaluate $y\ln(y)$ you would still need the use of Hospital to have a rigorous proof $\endgroup$ – DMH16 Mar 3 '17 at 5:26
  • $\begingroup$ Actually you don't need it (besides I never used this rule in my life). One way (probably not the faster one) is to say that $\ln(y)y = -x/e^x$ (with $x=-\ln(y)$), then you have to find the limit of $e^x/x$, when $x$ goes to $\infty$. This limit exists (real or $\infty$) since the function is non-decreasing for $x>1$. If it is a real limit, say $l>1$, then for all $x>2$, $e^x/x<=l$. Let $x_0>2$ s.t. $e^{x_0}=3lx_{0}/2$. Then $e^{2x_0}=9lx_{0}^2/4>=2l(2x_{0})$. So $e^{x_1}/x_1 > l$, with $x_1=2x_0$. That is a contradiction and then the limit should be $\infty$. Then $y\ln(y) \rightarrow 0$. $\endgroup$ – fonfonx Mar 3 '17 at 6:00
  • $\begingroup$ @DMH16: Another way (if your exponential function is defined by the series) is to simply note that $e^x > 1+x+\frac12 x^2$ for positive real $x$. A third way (if say you know that $e^x > 2^x$ for positive real $x$) is to observe that $e^{2x} / (2x) > (2^x/2) \times (e^x/x) > 2 (e^x/x)$ for integer $x > 2$ and that $e^x / x > e^y / (2y)$ where $y$ is the largest power of $2$ not greater than $x$, and finish by induction, giving a proof (of course based on the 'known' inequality) that does not use differentiation at all! $\endgroup$ – user21820 Mar 3 '17 at 9:02
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I would like to point out that your calculation is not quite as wrong as you might think. The point is that obviously you cannot replace any subexpression in side a limit with its limit, but there is the algebraic limit theorem (see e.g. wikipedia or this question) which allows certain such operations. For the case at hand, it states that if $\lim_{u\to\infty} f(u)$ and $\lim_{u\to\infty} g(u)$ exist and one of them is finite, then $$ \lim\limits_{u\to\infty} \!\left(f(u)+ g(u)\right)=\lim\limits_{u\to\infty} f(u) + \lim\limits_{u\to\infty} g(u)$$ and $$ \lim\limits_{u\to\infty} f(u)\cdot g(u)=\lim\limits_{u\to\infty} f(u) \cdot \lim\limits_{u\to\infty} g(u)\,.$$ For your case, we have $$\lim\limits_{u\to\infty} \!\left[-\frac{\ln\left(u^2\right)}{1+\frac1{u^2}}+\ln\left(1+u^2\right)\right]\\ =\lim\limits_{u\to\infty} \!\left[\left(-\ln u^2+\left(1+\frac{1}{u^2}\right)\ln\!\left(1+u^2\right)\right)\cdot\underbrace{\left(\frac{1}{1+\frac{1}{u^2}}\right)}_{g(u)}\right]$$ Now $\lim_{u\to\infty}g(u)=1$, so we can drop it by the product rule above. We are left with $$\lim\limits_{u\to\infty} \!\left[\ln\left(\frac{1+u^2}{u^2}\right)+\underbrace{\frac{\ln\!\left(1+u^2\right)}{u^2}}_{h(u)}\right]$$ Here, $\lim_{u\to\infty}h=0$, and we can use the sum rule above to discard it and finally arrive at your result.

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What you are essentially doing is to use the following claim: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

(Wrong) $\lim_{x \to p} f(g(x),h(x)) = \lim_{x \to p} f(\lim_{y \to p} g(y),h(x))$.

For a simple but instructive counter-example:

$1 = \lim_{x \to 0} \lfrac{\exp(2x)-(1+x)}{x} \ne \lim_{x \to 0} \lfrac{\exp(2x)-(1)}{x} = 2$ even though $\lim_{x \to 0} (1+x) = 1$.

Also, although I always advise using asymptotic expansion, we can 'avoid' using it in our final solution even though we can derive it via asymptotic expansion. Namely we look at the most significant terms in the expansions and pull them out by subtraction or division as appropriate, such that we can easily deal with the rest.

$\lfrac{\ln(x^2)}{1+x^2}-\ln\big(\lfrac{x^2}{1+x^2}\big) = \ln(x^2) - \lfrac{x^2 \ln(x^2)}{1+x^2} - \ln(x^2) + \ln(1+x^2) = - \lfrac{x^2 \ln(x^2)}{1+x^2} + \ln(1+x^2)$.

As $x \to 0$, we clearly have $\ln(1+x^2) \to 0$ and $x^2 \ln(x^2) = \lfrac{-\ln(1/x^2)}{1/x^2} \to 0$ since $1/x^2 \to \infty$. You can prove these by your favourite methods, two of which I have sketched in this comment.

Another issue in your question that I realize has not been addressed is that it is incorrect to let $x = \lfrac1u$. Think of it this way. You are given $x \to 0$, meaning that $x$ is going towards $0$ in some manner that you have no control over. All you know is that it eventually remains as close to $0$ as you desire, but never equal. You are free to let $u = \lfrac1x$, but then you cannot claim that $u \to \infty$. It may not since $x$ may oscillate about $0$. But you can let $u = \lfrac1{x^2}$ and claim that $u \to \infty$, because it is true and can be proven.

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I will add this modified version as another answer, since the last has been downvoted to oblivion. The important part of the answer is still using the continuity of the exponential to calculate the limit without L'Hôpital or series. We have that

\begin{align} &\exp\left(\lim_{x\to 0}\frac{\ln\left(x^2\right)}{1+x^2}-\ln\left(\frac{x^2}{1+x^2}\right)\right)\\ =&\lim_{x\to 0}\exp\left(\frac{\ln\left(x^2\right)}{1+x^2}-\ln\left(\frac{x^2}{1+x^2}\right)\right)\\ =&\lim_{x\to 0}{\left(x^2\right)}^{\frac{1}{1+x^2}}\cdot\frac{1+x^2}{x^2} \end{align}

Now, write $x^2=t$, then the limit becomes

$$\lim_{t\to0^+}t^{\frac1{1+t}}\cdot (1+t^{-1})=t^{\frac1{1+t}}+t^{\frac{-t}{1+t}}$$

The first summand very clearly tends $0$. Now, the second can be written as

$${\left(t^t\right)}^{-(1+t)}$$

and $t^t$ tends to $1$ as $t\to0^+$ (see the lemma below). Since $-(1+t)\to-1$, and in particular it is away from $0$ near the limit (see the bonus section), we have that ${\left(t^t\right)}^{-(1+t)}\to 1$. It follows that

$$\exp\left(\lim_{x\to 0}\frac{\ln\left(x^2\right)}{1+x^2}-\ln\left(\frac{x^2}{1+x^2}\right)\right)=1$$

So the limit inside must be $0$.


Lemma: $t^t\to1$ as $t\to0^+$

Proof (without using series or L'Hôpital): We will show that $\sqrt[n]{n}\to 1$ as $n\to\infty$. This implies that

$${\left(\frac{1}{n}\right)}^{1/n}\to1$$

as $1/n\to0^+$, which in turn proves the lemma via continuity of $t^t$, or via monotocinty near $t=0$, which can be concluded by differentiating $t^t=\exp\big(t\ln(t)\big)$.

This is a classical limit, but we can write a short proof here. We have that $\sqrt[n]{n}>1$ for all $n>1$. For each $n>1$, let $\epsilon_n>0$ be such that $\sqrt[n]{n}=1+\epsilon_n$. Then, by the binomial theorem:

$$n={(1+\epsilon_n)}^n=1+\binom{n}{1}\epsilon_n+\binom{n}{2}{\epsilon_n}^2+\dots$$

Since each term in the sum on the RHS is positive, we conclude that

$$n<\binom{n}{2}{\epsilon_n}^2=\frac{n(n-1)}{2}{\epsilon_n}^2,$$

and hence

$$0<\epsilon_n<\sqrt{\frac{2}{n-1}}.$$

The last term goes to $0$ as $n\to\infty$, so by the squeeze theorem, $\epsilon_n$ also goes to $0$ as $n\to\infty$. $\square$


This is a bit of a bonus I wrote down after my discussion with user21820. It's a very basic, real analysis $\epsilon$-$\delta$ style proof of the limit used in the answer. There is no particularly clever idea involved, but rather appropriate symbolic manipulation. It also sheds light into why $c>0$ is important in the statement below.

Claim: Suppose that, as $x\to 0$, $g(x)\to1$ and and $f(x)\to c$ for some $c>0$. Then

$$\lim_{x\to0}{g(x)}^{f(x)}=1$$

Proof: By hypothesis, the following hold:

\begin{align} \forall\epsilon_f>0,\,\exists\delta_f>0,\,\forall x\,\text{ with }\,0<|x|<\delta_f,\,|f(x)-c|<\epsilon_f \tag{1}\label{eq1}\\ \forall\epsilon_g>0,\,\exists\delta_g>0,\,\forall x\,\text{ with }\,0<|x|>\delta_g,\,|g(x)-1|<\epsilon_g\tag{2}\label{eq2} \end{align}

We will show that

\begin{align} \forall\epsilon>0,\,\exists\delta>0,\,\forall x\,\text{ with }\,0<|x|<\delta,\,\left|{g(x)}^{f(x)}-1\right|<\epsilon \tag{$*$}\label{eq*} \end{align}

Let $\epsilon>0$ and choose $0<\epsilon_f<c$ for $\eqref{eq1}$. Then there is some $\delta_f>0$ such that whenever $0<|x|<\delta_f$ we have $0<c-\epsilon_f<f(x)<c+\epsilon_f$. In this case, it holds that

$$\tag{3}\label{eq3}0<\frac1{c+\epsilon_f}<\frac1{f(x)}<\frac1{c-\epsilon_f}.$$

Now, we have that

$$k_0={(1+\epsilon)}^{\displaystyle\left(\frac1{c+\epsilon_f}\right)}-1>0$$

so we may take $\epsilon_g=k_0$ in $\eqref{eq2}$ to obtain some $\delta_0>0$ such that whenever $0<|x|<\delta_0$ we have

$$g(x)<1+k_0={(1+\epsilon)}^{\displaystyle\left(\frac1{c+\epsilon_f}\right)}\tag{4}\label{eq4}$$

Similarly, we have that

$$k_1=1-{(1-\epsilon)}^{\displaystyle\left(\frac1{c+\epsilon_f}\right)}>0$$

so we may take $\epsilon_g=k_1$ in $\eqref{eq2}$ to obtain some $\delta_1>0$ such that whenever $0<|x|<\delta_1$ we have

$$1-k_1={(1-\epsilon)}^{\displaystyle\left(\frac1{c+\epsilon_f}\right)}<g(x)\tag{5}\label{eq5}$$

Now, let $\delta=\min\{\delta_f,\delta_0,\delta_1\}>0$. Then, whenever $0<|x|<\delta$, we have from $\eqref{eq4}$ and $\eqref{eq5}$ that

$${(1-\epsilon)}^{\displaystyle\left(\frac1{c+\epsilon_f}\right)}<g(x)<{(1+\epsilon)}^{\displaystyle\left(\frac1{c+\epsilon_f}\right)}\tag{6}\label{eq6}$$

From our choice of $\delta_f$, it follows from $\eqref{eq3}$ that whenever $0<|x|<\delta$

$${(1+\epsilon)}^{\displaystyle\left(\frac1{c+\epsilon_f}\right)}<{(1+\epsilon)}^{1/f(x)}\\ {(1-\epsilon)}^{1/f(x)}<{(1-\epsilon)}^{\displaystyle\left(\frac1{c+\epsilon_f}\right)}$$

The first inequality follows from the fact that $h(x)=a^x$ is increasing when $a>1$ and the second follows from the fact that it is decreasing when $a<1$. The inequalities above, together with $\eqref{eq6}$, imply that whenever $0<|x|<\delta$

$${(1-\epsilon)}^{1/f(x)}<g(x)<{(1+\epsilon)}^{1/f(x)}$$

Since $f(x)>0$ in the inequality above , it follows that $1-\epsilon<{g(x)}^{f(x)}<1+\epsilon$, that is, $|{g(x)}^{f(x)}-1|<\epsilon$, which proves $\eqref{eq*}$ and thus the claim. $\square$

Notice that in this last step, we needed $f(x)>0$ in order to preserve the direction of the inequalities. That's where $c>0$ comes into play.

Of course, adapting it to $c<0$ is simple, but for $c=0$ the claim does not extend. Consider $g(x)=e^{1/x^2}$ and $f(x)=x^2$.

This whole discussion can be simplified by noticing that $x^y$ is continuous in $(0,\infty)^2$. Now, when $x=0$ and $y>0$, $x^y=0$. But when $x>0$ and $y=0$, $x^y=1$, so already we see there's some funky discontinuity going on if we try to extend the domain to $[0,\infty)$.

Near $(0,0)$, things get even weirder, and you can toy with the previous $g(x)$-$f(x)$ example to show that different paths can lead you to any non-negative limit.

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  • $\begingroup$ Still incorrect; just because $t^t \to 1$ as $t \to 0^+$ does not imply that $(t^t)^{f(t)} \to 1^{\lim_{u \to 0^+} f(u)}$ as $t \to 0^+$. Read this way, you're making exactly the very kind of error I pointed out in my post. Note that invalidity of a deduction does not require a counter-example (due to incompleteness theorems). If you're using a specific theorem to handle this case, state it clearly as it is very misleading to students otherwise. $\endgroup$ – user21820 Mar 3 '17 at 17:26
  • $\begingroup$ You misunderstand, because $f(t)\to-1$ in this case. Or, if you prefer, consider $$\frac{1}{{(t^t)}^{f(t)}},$$ and both $t^t$ and $f(t)$ tend to $1$. No theoerem needed here. $\endgroup$ – Fimpellizieri Mar 3 '17 at 18:40
  • $\begingroup$ @user21820 Your comment really bothers me because it delegitimizes a correct answer. $\endgroup$ – Fimpellizieri Mar 3 '17 at 21:37
  • $\begingroup$ When I say "incorrect" I do not mean "false". Another kind of answer that I label as "incorrect" is a 'proof' that only asserts true statements throughout but in a logically incoherent way. This commonly occurs in wrongly switching quantifiers, and yet no one can provide a counter-example to any of its statements. In your case, you are not using the usual notion of continuity, where for any continuous $f$ you have that $\lim_{x \to a} f(g(x)) = f( \lim_{x \to a} g(x) )$ if $\lim_{x \to a} g(x)$ is a real number. That is precisely why I said you must state the theorem you use clearly. $\endgroup$ – user21820 Mar 4 '17 at 2:56
  • $\begingroup$ I fail to understand what you label as 'incorrect' here. There is no switch of quantifiers here. $\endgroup$ – Fimpellizieri Mar 4 '17 at 2:58

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