Find without using De L'Hospital's rule the following limit:
$$\lim_\limits{x\to +\infty}{\left( \left( e+1\right) ^{\ln \left( e^x+1\right)} - \left( e+1\right) ^x\right)} $$
I have tried to factorize it but I always seem to end up with an indeterminate form... How can I do it with using DLH?
Please don't use approximations because I haven't "officially" learnt them yet...
If no L'Hopital and asymptotics are allowed then we need some kind of "standard" limits to be used. For example, if we know the derivatives of $\ln x$ and $a^x$ at zero then we can write by the derivative definition \begin{align} &\frac{\ln(1+h)}{h}\to 1,\tag{1}\\ &\frac{a^h-1}{h}\to\ln a\tag{2} \end{align} as $h\to 0$.
Now denote $a=e+1$ and factor out $a^x$ to get $$ \Bigl(a^{\ln(1+e^{-x})}-1\Bigr)a^x=\underbrace{\frac{a^{\ln(1+e^{-x})}-1}{\ln(1+e^{-x})}}_{\to\ln a\ \text{ (by (2))}}\cdot \underbrace{\frac{\ln(1+e^{-x})}{e^{-x}}}_{\to 1\ \text{ (by (1))}}\cdot \underbrace{e^{-x}a^x}_{\to\infty}\to\infty. $$