$$\lim_{x \rightarrow +\infty} \frac{2^x}{x}$$ $$\lim_{x \rightarrow \infty} \frac{x^{50}}{e^x}$$
I don't really know how to solve this.
As for the first one, I know that $\lim_{x \rightarrow \infty} a^x=0$ , I supposed that helps...?
How do I solve these (preferably analytically, but I'll also accept otherwise)?
Hint for the first.
taking logarithm we get
$$\lim_{x\to +\infty}(x\ln(2)-\ln(x))=$$
$$\lim_{x\to +\infty} x\left(\ln(2)-\frac{\ln(x)}{x}\right)=$$
$$+\infty$$
since $\lim_{x\to+\infty}\frac{\ln(x)}{x}=0$.
thus the first limit is $+\infty$.
the same approach gives $0$ for the second.
You say that $\lim_{x\to\infty} a^x = 0$, however this is only true if $|a| < 1$, and hence not true for $2^x$ or $e^x$.
With that said, you can solve both of these limits if you know that exponentials eventually dominate any polynomial (if the base of the exponential is larger than 1). Let $p$ be a polynomial and $a>1$, then $$\lim\limits_{x\to\infty} \frac{p(x)}{a^x} = 0 \quad\quad \lim\limits_{x\to\infty} \frac{a^x}{p(x)} = \pm \infty,$$ where the sign in the last limit is the same as the sign of the term with the highest degree in the polynomial.
You can now solve the limits, remembering that both $2^x$ and $e^x$ are exponentials with base larger than 1, i.e. $2>1$ and $e>1$, and $x$ and $x^{50}$ are polynomials.
Hint: $$2^x=\sum_{k=0}^{\infty }\frac{(x\log 2)^k}{k!}$$ and $$\frac{x^{50}}{e^x}=\frac{50!x^{50}}{x^{50}+50!(1+x+\frac{x^2}{21}+....\frac{x^{49}}{49!}+\frac{x^{51}}{51!}......)}$$
You can catch two birds with a stone. Note that $2^x=e^{x\log 2}$ so $$ \frac{2^x}{x}=\frac{e^{x\log 2}}{x}=\frac{e^{x\log 2}}{x\log 2}\log 2 =\frac{e^t}{t}\log 2 \qquad(\text{for }t=x\log2) $$ and $$ \frac{x^{50}}{e^x}=\left(\frac{x}{e^{x/50}}\right)^{\!50}= 50^{50}\left(\frac{x/50}{e^{x/50}}\right)^{\!50}= 50^{50}\left(\frac{t}{e^t}\right)^{\!50}\qquad(\text{for }t=x/50) $$ Thus you only have to compute $$ \lim_{t\to\infty}\frac{e^t}{t} $$
