${\lim_{x\rightarrow \infty}}(\sqrt{x^2+2x+3} - \sqrt{x^2+3})^{x}$ $${\lim_{x\rightarrow \infty}}(\sqrt{x^2+2x+3} - \sqrt{x^2+3})^{x}$$
I tried taking log both sides (on paper). 
After taking log how do I proceed? You get $\infty$ * $(\infty-\infty$). But $\infty-\infty$ could be any number. How can I take this as 0? Even if I take it as zero, the question becomes nasty. 
Please help me solve this limit. It must be solved without using series expansion as this isn't taught. Only L'Hopital and basic algebra. 

$$A=\left(\sqrt{x^2+2x+3} - \sqrt{x^2+3}\right)^{x}$$ For what is inside the parenthesis, use Taylor
$$\sqrt{x^2+2x+3}=x+1+\frac{1}{x}-\frac{1}{x^2}+O\left(\frac{1}{x^3}\right)$$
$$\sqrt{x^2+3}=x+\frac{3}{2 x}+O\left(\frac{1}{x^3}\right)$$
$$\sqrt{x^2+2x+3} - \sqrt{x^2+3}=1-\frac{1}{2 x}-\frac{1}{x^2}+O\left(\frac{1}{x^3}\right)$$
$$\log(A)=x \log\left(1-\frac{1}{2 x}-\frac{1}{x^2}+O\left(\frac{1}{x^3}\right) \right)=-\frac{1}{2}-\frac{9}{8 x}+O\left(\frac{1}{x^2}\right)$$
$$A=e^{\log(A)}=\frac{1}{\sqrt{e}}-\frac{9}{8 \sqrt{e} x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.
