Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ [duplicate]

Solution 1:

Your result is correct. A simpler method is to decompose this limit into a product of simpler limits:

Let $f(x)=(1+x)^{1/x}$. We have $\lim_{x\rightarrow 0} f(x) = e$. Note that your expression can be written as $$ \lim_{x\rightarrow 0} \frac{e^{f(x)}- f(x)^e}{x^2} = \lim_{x\rightarrow 0} \frac{e^{f(x)}- f(x)^e}{(f(x)-e)^2}\frac{(f(x)-e)^2}{x^2} = \lim_{y\rightarrow e} \frac{e^y - y^e}{(y-e)^2} \cdot \left(\lim_{x\rightarrow 0} \frac{f(x)-e}{x}\right)^2 $$

From l'Hôpital's rule we have $$ \lim_{y\rightarrow e} \frac{e^y - y^e}{(y-e)^2} =^H \lim_{y\rightarrow e} \frac{e^y - ey^{e-1}}{2(y-e)} =^H \lim_{y\rightarrow e} \frac{e^y - e(e-1)y^{e-2}}{2} = \frac12 e^{e-1}$$

$$ \lim_{x\rightarrow 0} \frac{f(x)-e}{x} = \lim_{x\rightarrow 0} \frac{e^{\ln(1+x)/x}- e}{\ln(1+x)/x - 1}\frac{\ln(1+x)/x - 1}{x} = \lim_{y\rightarrow 1}\frac{e^y -e}{y-1} \lim_{x\rightarrow 0}\frac{\ln(1+x)-x}{x^2} $$ $$ \lim_{y\rightarrow 1}\frac{e^y -e}{y-1} =^H \lim_{y\rightarrow 1} \frac{e^y}{1} = e $$ $$ \lim_{x\rightarrow 0}\frac{\ln(1+x)-x}{x^2} =^H \lim_{x\rightarrow 0}\frac{\frac{1}{1+x}-1}{2x} = \lim_{x\rightarrow 0}\frac{-1}{2(1+x)} = -\frac12$$

In total $$ \lim_{x\rightarrow 0} \frac{e^{f(x)}- f(x)^e}{x^2} = \frac12 e^{e-1} \cdot\left(-\frac{e}{2}\right)^2 = \frac{e^{e+1}}{8}$$

Solution 2:

Looks good, and thumbs up for taking the time to provide your own work!

$$\lim\limits_{x \to 0}\frac{e^{(1+x)^{1/x}}-(1+x)^{e/x}}{x^2}$$

Not sure if this is any better, but taking $t=\left(1+x\right)^{1/x}$ turns the numerator into $e^t-t^e$ with $t \to e$ when $x \to 0$. Its first non-zero term in the series expansion (around $t=e$) is: $$\frac{1}{2} e^{e - 1} \left(t - e\right)^2 \tag{$\star$}$$ Now borrowing your expansion of $t=\left(1+x\right)^{1/x}$ around $x=0$:

$$(1+x)^{\frac{1}{x}}=\color{blue}{e-\frac{ex}{2}}+\frac{11}{24}ex^2+\ldots$$

And plugging this only up to order $1$ (blue) into $(\star)$ gives: $$\frac{1}{2} e^{e - 1} \left(\color{blue}{e-\frac{ex}{2}} - e\right)^2 =\frac{1}{8} e^{e + 1} x^2$$

For $(\star)$, with $f(t)=e^t-t^e$ you have:

• $f'(t)=e^t-et^{e-1} \implies f'(e)=0$
• $f''(t)=e^t-e(e-1)t^{e-2} \implies f''(e)=e^{e-1}$

So:

$$\begin{align}f(t) & = f(e)+f'(e)(t-e)+\frac{1}{2}f''(e)(t-e)^2 + \ldots \\ & = 0 + 0 + \frac{1}{2} e^{e - 1} \left(t - e\right)^2 + \ldots\end{align}$$

Solution 3:

Another Solution

\begin{align*} &\lim_{x \to 0+}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}\\ =&\lim_{x \to 0+}\frac{e^{(1+x)^{\frac{1}{x}}}-e^{\frac{e}{x}\ln(1+x)}}{x^2}\\ =&\lim_{x \to 0+}\left[e^{\frac{e}{x}\ln(1+x)}\cdot\frac{e^{(1+x)^{\frac{1}{x}}-\frac{e}{x}\ln(1+x)}-1}{x^2}\right]\\ =&e^e\cdot \lim_{x \to 0+}\frac{e^{(1+x)^{\frac{1}{x}}-\frac{e}{x}\ln(1+x)}-1}{x^2}\\ =&e^e\cdot \lim_{x \to 0+}\frac{(1+x)^{\frac{1}{x}}-\frac{e}{x}\ln(1+x)}{x^2}\\ =&e^e\cdot \lim_{x \to 0+}\frac{(e-\frac{1}{2}ex+\frac{11}{24}ex^2+\cdots)-\frac{e}{x}(x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots)}{x^2}\\ =&e^e\cdot \frac{1}{8}e\\ =&\frac{e^{e+1}}{8}. \end{align*}

Solution 4:

In response to reading some of the comments, I just want to say that the L'hospital calculation is not horrible, if one works a little strategically. Of course, this is not entriely tedium free...

Let $$g(x) = \frac{\log(1+x)}{x}.$$ We have $g \to 1,$ and $g' \to -1/2$ as $x \to 0.$ (you'll have to show this for $g'$ but this is not hard).

Now let the limit in the question be written as $$L = \lim \frac{f_1(x) - f_2(x)}{x^2}, $$ where

\begin{align} f_1(x) &:= e^{e^{g(x)}} \\ f_2(x) &:= e^{eg(x)} \end{align}

(I'll now start dropping the $(x)$ in $g(x)$ for legibility.)

By the chain rule, \begin{align} f_1' &= \left(e^{g(x)}\right)' e^{e^{g}} = g' e^{g} e^{e^{g}} \\ f_2' &= eg' e^{eg}\end{align}

So, by L'Hospital, $$ L = \lim \frac{g'e^g \left(e^{e^g} -e e^{(e-1)g }\right) }{2x}$$

But, $g'e^{g} \to -e/2$, so $$ L = -\frac{e}{4} \lim \frac{e^{e^g} - e^{(e-1)g + 1} }{x} $$ (assuming of course that the latter exists - this should be easy to show). The point of doing this is get back to something that looks (almost) like $f_1 - f_2$ on the numerator. In particular, I don't want the second derivative of $f_1$ to enter my expressions - that's too much work.

Applying L'Hospital to this again, we get $$ -\frac{4L}{e} = \lim g'e^g e^{e^g} - (e-1)g' e^{(e-1)g + 1} = -\frac{1}{2}e\cdot e^e + \frac{(e-1)}{2}e^e = -\frac{1}{2}e^e,$$ where I've simply evaluated using $g \to 1, g' \to -1/2,$ giving again $$ L = e \cdot e^e/8.$$

Solution 5:

Since both terms, say $A, B$, in numerator tend to same limit $e^e$ we can write $$A-B=B\cdot\frac{\exp (\log A - \log B) - 1}{\log A - \log B} \cdot(\log A - \log B) $$ and thus the numerator can be replaced by $e^e(\log A - \log B) $ or $$e^e\left((1+x)^{1/x}-e\cdot\frac{\log (1+x)}{x}\right)$$ Applying the same technique the above expression can be replaced by $$e^e\cdot e\left(\frac{\log(1+x)}{x}-1-\log\frac{\log(1+x)}{x}\right)$$ or $$e^{e+1}\cdot(u-\log(1+u))$$ where $$u=\frac{\log(1+x)}{x}-1\to 0$$ On the other hand we also know via L'Hospital's Rule or Taylor series that $u/x\to - 1/2$ and therefore the expression $$\frac{u-\log(1+u)}{x^2}=\frac{u-\log(1+u)}{u^2}\cdot\frac{u^2}{x^2}$$ tends to $(1/2)(-1/2)^2=1/8$. The desired limit is thus $e^{e+1}/8$.

In general one should avoid multiplication/division and composition of Taylor series and use the famous ones directly from memory. Often the use of algebraic manipulation combined with standard limits reduces / alleviates the need of any gymnastics with Taylor series.