Find a limit, no Taylor formula 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.
 A: 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 $a.$
A: 
PRIMER:
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 $x<1$
and 
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag 2$$
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

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

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