Evaluate $$I=\lim_{x\to 0}\dfrac{(1+x)^{\frac{1}{x}}-(1+2x)^{\frac{1}{2x}}}{x}$$
My try: Use L'Hôpital's rule. We have $$I=\lim_{x\to0}(1+x)^{1/x}\left(\dfrac{x}{x+1}-\ln{(x+1)}\right)-(1+2x)^{1/2x}\left(\dfrac{4x}{1+2x}-2\ln{(1+2x)}\right)$$
I'm stuck. Thank you very much for you help
Let $f(x)=(1+x)^{1/x}$. You know $f(0)=e$ makes $f$ continuous at $x=0$. Then you're looking at $$I=\lim_{x\to 0}\dfrac{ f(x)-f(0)-(f(2x)-f(0))}{x}$$
Thus, all you need to find is the derivative of $f(x)$ and $f(2x)$ at $x=0$.
Hint:
Applying Taylor's expansion
$$(1+x)^{\frac{1}{x}}=\exp\left\{\frac{\ln(1+x)}{x}\right\}=\exp\left\{1-\frac{x}{2}+O(x)\right\}=e-\frac{ex}{2}+O(x)$$
$$(1+2x)^{\frac{1}{2x}}=\exp\left\{\frac{\ln(1+2x)}{2x}\right\}=\exp\left\{1-x+O(x)\right\}=e-ex+O(x)$$
Hence $$\lim_{x\to 0} \frac{(1+x)^{\frac{1}{x}}-(1+2x)^{\frac{1}{2x}}}{x}=\lim_{x\to 0}\left(\frac{e}{2}+\frac{O(x)}{x}\right)=\frac{e}{2}$$
$$\lim_{x\to 0} \frac{(1+x)^{\frac{1}{x}}-(1+2x)^{\frac{1}{2x}}}{x}=-\lim_{x\to 0} \frac{f(x)-f(2x)}{x-2x}=-\lim_{x\to 0} f'(x)=\lim_{x\to 0} \left((1+x)^{\frac{1}{x}}\frac{\ln(1+x)-\frac{x}{1+x}}{x^2}\right)=e\times \frac{1}{2}=\frac{e}{2}$$
– Iloveyou Jan 20 '14 at 16:23