Here is problem:

$$\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x$$

The solution I presented in the picture below was made by a Mathematics Teacher

enter image description here

I tried to solve this Limit without using derivative (L'hospital) and Big O notation. Although I get the answer, I don't know if the technique I'm using definitely correct.

And here is my method:

$$\begin{align*}\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x&=\lim_{x \to \infty} \left(\frac {2x}{\sqrt{x^2+2x+3} +\sqrt{x^2+3}}\right)^x\\&=\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}\end{align*}$$

Then, I define a new function here

$$y(x)=\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1$$

We have

$$\begin{align*} \lim _{x\to\infty} y(x)&=\lim_{x \to \infty}\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1\\ &=\lim_{x \to \infty}(\sqrt{x^2+2x+3}-(x+1))+(\sqrt{x^2+3}-x)\\ &=\lim_{x \to \infty}\frac{2}{\sqrt{x^2+2x+3}+x+1}+ \lim_{x \to \infty}\frac{3}{\sqrt{x^2+3}+x}\\ &=0. \end{align*}$$

This implies that $$\lim_{x \to \infty}\frac{2x}{y(x)+1}=\infty $$


$$\begin{align*} \lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}&=\lim_{x \to\infty} \frac{1}{ \left(\frac{y(x)+2x+1}{2x} \right)^x}\\ &=\lim_{x \to\infty} \frac{1}{ \left(1+\frac{y(x)+1}{2x} \right)^x}\\ &=\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}\\ & \end{align*}$$

Here, we define two functions: $$f(x)=\left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}},\quad g(x)=\frac{y(x)+1}{2}. $$

We deduce that, $$ \lim_{x\to\infty} f(x)=e>0,\quad \lim_{x\to\infty} g(x)=\frac 12>0. $$ Thus, the limit $\lim_{x\to\infty} f(x)^{g(x)} $ exists and is finite.

Finally we get,

$$\begin{align*} \lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}} &=\frac{1}{\lim_{x \to \infty}\left( \left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}\right)}\\ &=\frac{1}{\left(\lim_{x\to\infty} \left( 1+\frac{1}{\frac{2x}{y(x)+1}} \right)^{\frac{2x}{y(x)+1}}\right)^{ \lim_{x\to\infty} \frac{y(x)+1}{2}}}\\ &=\frac {1}{e^{\frac12}}=\frac{\sqrt e}{e}.\\&& \end{align*}$$

Is the method I use correct?

I have received criticisms against my work. What can I do to make the method I use, rigorous? What are the points I missed in the method?

Thank you!

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    $\begingroup$ @DMcMor Thank you for edit my bad $\LaTeX$ $\endgroup$ – lone student Apr 12 '19 at 22:10
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    $\begingroup$ Honestly your TeX was pretty good! I just aligned the equations to make it a bit easier to see. $\endgroup$ – DMcMor Apr 12 '19 at 22:13
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    $\begingroup$ I'm also a BlackPenRedPen fan! (or is it BlackPenRedPenBluePen?) $\endgroup$ – Toby Mak Apr 13 '19 at 10:57
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    $\begingroup$ $x^y$ is continuous near $\left(e,\frac12\right)$; thus, $\lim\limits_{x\to\infty}f(x)^{g(x)}=\lim\limits_{x\to\infty}f(x)^{\lim\limits_{x\to\infty}g(x)}$ when $\lim\limits_{x\to\infty}f(x)=e$ and $\lim\limits_{x\to\infty}g(x)=\frac12$ $\endgroup$ – robjohn Apr 24 '19 at 9:59
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    $\begingroup$ No. By using the fact that $x^y$ is continuous at this point, it makes your argument rigorous. $\endgroup$ – robjohn Apr 24 '19 at 10:19

Your math looks good! I'd maybe just an extra step here and there to make it clear what your doing. Things like showing that you're multiplying by conjugates and maybe a change of variables, say $$z = \frac{2x}{y(x)+1},$$ near the end so it's a bit clearer where the $e$ comes from. Otherwise everything looks good! This is a tricky limit, I really like your solution.


The solution appears to be correct. Just for the sake of sanity, here's a different argument, based on the idea that knowing derivatives is knowing many limits.

First, find the limit of the logarithm of the beast, which is best treated also with the substitution $x=1/t$, which makes us trying to find $$ \lim_{t\to0^+}\frac{1}{t}\log\left(\frac{\sqrt{1+2t+3t^2}-\sqrt{1+3t^2}}{t}\right) = \lim_{t\to0^+}\frac{1}{t}\log\left(\frac{2}{\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}}\right) $$ This can be rewritten as $$ \lim_{t\to0^+}-\frac{\log\bigl(\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}\,\bigr)-\log2}{t} $$ which is the negative of the derivative at $0$ of $$ f(t)=\log\bigl(\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}\,\bigr) $$ Since $$ f'(t)=\frac{1}{\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}}\left(\frac{1+3t}{\sqrt{1+2t+3t^2}}+\frac{3t}{\sqrt{1+3t^2}}\right) $$ we have $f'(0)=1/2$ and therefore the limit is $-1/2$, so your given limit $$ e^{-1/2} $$


Your approach is correct but its presentation / application is more complicated than needed here.

Here is how you can use the same approach with much less effort. You have already observed that the base $$F(x) =\sqrt {x^2+2x+3}-\sqrt{x^2+3}$$ tends to $1$ as $x\to\infty $. Now the expression under limit can be written as $$\{F(x) \} ^x=\{\{1+(F(x)-1)\}^{1/(F(x)-1)}\}^{x(F(x)-1)}$$ The inner expression tends to $e$ and the exponent $x(F(x) - 1)\to -1/2$ so that the desired limit is $e^{-1/2}$.

Another part of your approach is that it involves the tricky use of subtracting $2x+1$ from $y(x) $. For those who are experienced in the art of calculus this step is obvious via the approximation $$\sqrt{x^2+2ax+b}\approx x+a$$ but it may appear a bit mysterious for a novice. It is best to either explain this part or remove it altogether as I have done it in my answer.

Also note that your approach uses the following limits / rules (it is not necessary to point them out explicitly unless demanded by some strict examiner) :

  • $\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x=e$
  • If $\lim_{x\to\infty} f(x) =a>0$ and $\lim_{x\to\infty} g(x) =b$ then $\{f(x) \} ^{g(x)} \to a^b$ as $x\to\infty $.

By squaring, we can verify that $$ x\le\sqrt{x^2+3}\le x\left(1+\frac3{2x^2}\right)\tag1 $$ and $$ x+1\le\sqrt{x^2+2x+3}\le(x+1)\left(1+\frac1{x(x+1)}\right)\tag2 $$ Adding $(1)$ and $(2)$ gives $$ 2x+1\le\sqrt{x^2+2x+3}+\sqrt{x^2+3}\le(2x+1)\left(1+\frac3{2x^2}\right)\tag3 $$ Multiplying numerator and denominator by $\sqrt{x^2+2x+3}+\sqrt{x^2+3}$ gives $$ \sqrt{x^2+2x+3}-\sqrt{x^2+3}=\frac{2x}{\sqrt{x^2+2x+3}+\sqrt{x^2+3}}\tag4 $$ Bernoulli and cross multiplying yield $$ 1-\frac3{2x}\le\left(1-\frac3{2x^2}\right)^x\le\left(1+\frac3{2x^2}\right)^{-x}\tag5 $$ Therefore $(3)$, $(4)$, and $(5)$ yield $$ \left(\frac{2x}{2x+1}\right)^x\left(1-\frac3{2x}\right)\le\left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x\le\left(\frac{2x}{2x+1}\right)^x\tag6 $$ The Squeeze Theorem then says $$ \lim_{x\to\infty}\left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x=e^{-1/2}\tag7 $$


Your method is correct if and only if you truly understand how to rigorously prove a critical step where you effectively claim $\lim_{x∈\mathbb{R}→∞} (1+\frac1x)^{f(x)} = \lim_{x∈\mathbb{R}→∞} e^{f(x)/x}$. Note that this requires real exponentiation, and the simplest proof of this would involve the asymptotic expansions for $\exp,\ln$, hence I personally think it's misleading to think of your method as successfully evading asymptotic expansions. To preempt the very common bogus proof, note that this claim does not follow from $\lim_{x∈\mathbb{R}→∞} (1+\frac1x)^x = e$.

After I first posted my answer, you edited your attempt in a non-trivial way. (Please do not edit your question in this way in the future, as it invalidates existing answers.) It still has the same conceptual error, just with different appearance. In this second attempt, you effectively claim that if $\lim_{x∈\mathbb{R}→∞} g(x) = c$ then $\lim_{x∈\mathbb{R}→∞} f(x)^{g(x)} = \lim_{x∈\mathbb{R}→∞} f(x)^c$ if the latter limit exists. This is not in general true! If you can state and prove the correct theorem of this sort (in a comment), then I will believe that you understand it.

The common underlying error is that you replaced part of a limit expression with its limit, which is in general invalid!

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – quid Apr 13 '19 at 10:41
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    $\begingroup$ "The common underlying error is that you replaced part of a limit expression with its limit, which is in general invalid!" Definitely Wrong argument. Because, all limit rules are "invalid generally". You cannot show the applicable limit rule in all cases. $\endgroup$ – Zaharyas Apr 20 '19 at 11:28
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    $\begingroup$ Your last sentence is mathematically non sense. (as it appears). $\endgroup$ – Zaharyas Apr 20 '19 at 11:45
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    $\begingroup$ I'm IMO gold medalist. Now, I study bachelors. $\endgroup$ – Zaharyas Apr 20 '19 at 11:52
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    $\begingroup$ IMO does not contain calculus or limit theorems. The content ranges from extremely difficult algebra and pre-calculus problems to problems on branches of mathematics not conventionally covered at school and often not at university level either, such as projective and complex geometry, functional equations, combinatorics, and well-grounded number theory, of which extensive knowledge of theorems is required. You asked me about my mathematical background. I gave the answer. $\endgroup$ – Zaharyas Apr 20 '19 at 18:46

The proof can be accelerated, using binomial Maclaurin series in the form of $$\sqrt{x^2+2x+3} = (x+1)\sqrt{1+\dfrac2{(x+1)^2}} = (x+1)\left(1 + \dfrac1{(x+1)^2}+O\left(x^{-4}\right)\right)$$ $$ = x+1+\dfrac1x+O\left(x^{-2}\right),$$ $$\sqrt{x^2+3} = x\left(1+\dfrac3{2x^2}+O(x^{-4})\right) = x + \dfrac3{2x}+O(x^{-3}).$$ Then $$\ln L = \ln \lim\limits_{x\to\infty} \left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x = \lim\limits_{x\to\infty} x\ln\left(1-\frac1{2x}+O\left(x^{-2}\right)\right)$$ $$= \lim\limits_{x\to\infty} x\left(-\frac1{2x}+O\left(x^{-2}\right)\right) = -\frac12,$$ $$L=e^{\Large^{-\frac12}}.$$


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