I want to calculate the limit $\displaystyle{\lim_{x\rightarrow \infty}\frac{2x^{1000}}{e^{2x}+1}}$.
Since we get the undefinite form $\frac{\infty}{\infty}$ we use the De L'Hospital Rule and we get
\begin{equation*}\lim_{x\rightarrow \infty}\frac{2x^{1000}}{e^{2x}+1}=\lim_{x\rightarrow \infty}\frac{\left (2x^{1000}\right )'}{\left (e^{2x}+1\right )'}=\lim_{x\rightarrow \infty}\frac{2\cdot 1000x^{999}}{2e^{2x}}=\lim_{x\rightarrow \infty}\frac{1000x^{999}}{e^{2x}}\end{equation*}
We apply the DLH again and we get \begin{equation*}\lim_{x\rightarrow \infty}\frac{1000x^{999}}{e^{2x}}=\lim_{x\rightarrow \infty}\frac{\left (1000x^{999}\right )'}{\left (e^{2x}\right )'}=\lim_{x\rightarrow \infty}\frac{1000\cdot 999x^{998}}{2e^{2x}}\end{equation*}
We see that we have to apply the DLH other $998$ times and then we get: \begin{equation*}\lim_{x\rightarrow \infty}\frac{1000\cdot 999\cdot 998\cdot \ldots \cdot 2\cdot 1x^{0}}{2^{999}e^{2x}}=\lim_{x\rightarrow \infty}\frac{1000\cdot 999\cdot 998\cdot \ldots \cdot 2\cdot 1}{2^{999}e^{2x}}=\lim_{x\rightarrow \infty}\frac{1000!}{2^{999}e^{2x}}=0\end{equation*}
Is this correct?
Could we calculate this limit also without the De L'Hospital Rule?