How to find this limit: $$ \lim_{x \to +\infty} \left[ (x+a)^{1+{1\over x}}-x^{1+{1\over x+a}}\right] $$ We know L'Hopital's rule, but don't know Taylor's formula.

  • 2
    $\begingroup$ Rewrite using standard tricks like $z=\ln(\exp(z))$ and proceed to simplify. $\endgroup$
    – Alex R.
    Nov 28, 2016 at 22:32
  • $\begingroup$ Basically you have to transform this into a form you can l'hopital. @AlexR suggested one way and the other way is to transform it into $lim \frac{1/f(x)-1/g(x)}{1/f(x)g(x)}$ but I think the exponential way will be easier to simplify. $\endgroup$
    – Zaros
    Nov 28, 2016 at 22:37
  • $\begingroup$ This is a hard one! Using MacLaurent series, you immediately see that the answer will be $a$, but getting it into a form to use l'Hospital's rule is not easy. One usual trick will be to exponentiate the expression; then you have a ratio (whose limit is $e^a$). But taking the derivatives is really messy, and you have to judiciously drop terms that "don't matter" in order to make progress. After a few derivatives and a clever trick the answer emerges, but it would be much easier to just derive the Taylor/MacLaurent theorems and use the series! $\endgroup$ Nov 28, 2016 at 23:46

5 Answers 5


Let's assume $a>0.$ The expression equals

$$\tag 1(x+a)^{1+1/x} - x^{1+1/x} + x^{1+1/x}- x^{1+1/(x+a)}.$$

Think of $x$ as fixed for the moment. Consider the function $f(y)= y^{1+x}.$ Then $f'(y) = (1+1/x)y^{1/x}.$ So by the mean value theorem,

$$\tag 2 (x+a)^{1+1/x} - x^{1+1/x} = (1+1/x)c^{1/x}\cdot a,$$

where $c \in (x,x+a).$ Since both $x^{1/x},(x+a)^{1/x} \to 1,$ we see $(2)\to a.$ That takes care of the first difference in $(1).$

For the second difference in $(1),$ think of $x $ fixed again and consider $g(y) = x^y.$ Then $g'(y) = \ln x \cdot x^y.$ So

$$\tag 3 x^{1+1/x}- x^{1+1/(x+a)}= \ln x\cdot x^c \cdot (1/x- 1/(x+a)),$$

where $c\in (1/x,1/(x+a)).$ From this you can conclude the second difference in $(1) \to 0.$ The desired limit is thus $a.$

  • $\begingroup$ This works since if the OP knows L'Hospital's Rule, then it is likely that the OP also knows the mean-value theorem. So, +1. I took a tedious and more elementary approach that uses only pre-calculus tools. $\endgroup$
    – Mark Viola
    Nov 28, 2016 at 23:50
  • $\begingroup$ Nice solution. Although it's interesting to find one without using MVT and inequalities, just L'Hopital and well-known limits. $\endgroup$ Nov 29, 2016 at 2:09
  • $\begingroup$ @AndreiPetrov But L'Hopital uses the MVT in its proof. $\endgroup$
    – zhw.
    Nov 29, 2016 at 19:10
  • $\begingroup$ I know, I meant, that I wanted to see a solution, more suitable to my taste, that is, the one using L'Hopital or well-known limits. I received it in another forum, see my own answer to this question. $\endgroup$ Nov 29, 2016 at 20:51
  • $\begingroup$ I should accept, that your method is quite useful, was unexpected to me and may be applied to other limit finding. So, I'm selecting it as the best at math exchange, although I personally prefer the one that came from outside. $\endgroup$ Nov 30, 2016 at 2:01


In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}} \tag 1$$

for $x<1$


$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag 2$$

for $x>0$.

Using elementary arithmetic, we can write

$$\begin{align} (x+a)^{1+{1/x}}-x^{1+{1/( x+a)}}&=x^{1/x}(x+a)\left(1+\frac ax\right)^{1/x}-x^{1/x}xx^{-\frac{a}{x(x+a)}}\\\\ &=ax^{1/x}\left(1+\frac ax\right)^{1/x}+x^{1/x}x\left(\left(1+\frac ax\right)^{1/x}-x^{-\frac{a}{x(x+a)}}\right) \tag 3 \end{align}$$

The limit as $x\to \infty$ of first-term on the right-hand side of $(3)$ is easily seen to be $a$. To see this, simply write $x^{1/x}=e^{\frac1x\log(x)}$ and note that $\log(x)/x \to 0$ as $x\to \infty$. Similarly, write $\left(1+\frac ax\right)^{1/x}=e^{\frac1x \log\left(1+\frac ax\right)}$ and note that $\frac1x \log\left(1+\frac ax\right)\to 0$ as $x\to \infty$.

The problem boils down to evaluating the second limit. Again writing

$$\left(1+\frac ax\right)^{1/x}-x^{-\frac{a}{x(x+1)}}=e^{\frac1x \log\left(1+\frac ax\right)}-e^{-\frac{a}{x(x+a)}\log(x)}$$

and using the inequalities in $(1)$ and $(2)$ we can show that

$$\frac{a}{x+a}+\frac{ax\log(x)}{x(x+a)+a\log(x)}\le x\left( \left(1+\frac ax\right)^{1/x}-x^{-\frac{a}{x(x+a)}}\right)\le \frac{ax}{x^2-a}+\frac{a\log(x)}{x+a} \tag 4$$

Applying the squeeze theorem to $(4)$ reveals that

$$\lim_{x\to \infty}x\left( \left(1+\frac ax\right)^{1/x}-x^{-\frac{a}{x(x+a)}}\right)=0$$

Putting everything together, we find that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty} (x+a)^{1+{1/x}}-x^{1+{1/( x+a)}}=a}$$

  • $\begingroup$ In (3) after the first equality second term is doubtful, it should be multiplied by x, I think. $\endgroup$ Nov 29, 2016 at 0:30
  • $\begingroup$ @AndreiPetrov That was a typographical error; I've edited. Nice catch. -Mark $\endgroup$
    – Mark Viola
    Nov 29, 2016 at 0:40
  • $\begingroup$ Please let me know how to improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$
    – Mark Viola
    Nov 29, 2016 at 0:46
  • $\begingroup$ Good coherent solution. Deserves the +1. $\endgroup$ Nov 29, 2016 at 1:51
  • $\begingroup$ @RajatMittal Thank you! Much appreciated. $\endgroup$
    – Mark Viola
    Nov 29, 2016 at 3:31

I think the best way to approach this one is to think about it conceptually. As $x$ approaches infinity, $a$ becomes insignificant in the exponent (make sure you see why). Also we know that $$\lim_{x→+∞}\frac{1}{x} = 0$$

Thus, the overall limit is simply

$$\lim_{x→+∞} ((x+a)-x) = a$$


$$x \to \infty \approx \left( {x + a} \right) \to \infty \Rightarrow {\left( {x + a} \right)^{1 + \frac{1}{x}}} \approx {x^{1 + \frac{1}{{x + a}}}} \Rightarrow \lim [{\left( {x + a} \right)^{1 + \frac{1}{x}}} - {x^{1 + \frac{1}{{x + a}}}}] = 0$$

  • $\begingroup$ This isn't kosher - $x +1 \approx x$ but $\lim_{x \to \infty} [(x+1) - x] = 1$. $\endgroup$ Nov 28, 2016 at 23:01
  • $\begingroup$ Unfortunately, this development completely lacks rigor. $\endgroup$
    – Mark Viola
    Nov 29, 2016 at 0:44

I received the best answer on a Russian forum. We don't need L'Hopital here. Just well-known limits. First, we take out a common factor $x^{1+{1\over x}}$, then subtract and add 1, use equation $z=e^{\ln z}$ for both terms and remember the limit $\lim_{x\to0}{e^x-1\over x}=1$ again for both terms. And for $x^{1\over x}$ we remember limit $\lim_{x\to+\infty}{\ln x\over x}=0$. As simple as that.


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