Let's suppose that $$\displaystyle\lim_{x \to{+}\infty}{f(X)}=\pm \infty$$ and $$\displaystyle\lim_{x \to{+}\infty}{g(x)}=\pm \infty$$, then we have that $$ \displaystyle\lim_{x \to{+}\infty}{\frac{f(x)}{g(x)}}=\frac{\pm \infty}{\pm \infty}$$ , so that we have an indeterminate form.
To know the value of the limit we will have to look at the functional form of every function and find the term of highest order. Then, according to the position (above or below of the fraction) of this element of highest order we will determine whether the limit is infinite, zero, or in case of being in both sides of the fraction, the value to which the limit is converging.
Let's remember how the limits of the main functions are ordered: $$$\log_{r} x < < x^n << x^m << a^x << b^x << x^x \mbox{ where } n< m \mbox{ and } b > a > 0$$$
Let's see some examples:
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If $$b>a>0$$ then $$\displaystyle\lim_{x \to{+}\infty}{\frac{x^k+log_{r}+a^x}{b^x-a^x}}=\displaystyle\lim_{x \to{+}\infty}{\frac{a^x}{b^x}}=0$$
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$$\displaystyle\lim_{x \to{+}\infty}{\frac{x^2+3^4x^3-x^4}{2x-\ln x+ x^2}}=\displaystyle\lim_{x \to{+}\infty}{\frac{-x^4}{x^2}}=- \infty$$
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$$\displaystyle\lim_{x \to{+}\infty}{\frac{2e^x-\log_{2}x}{x+1-3e^x}}=\displaystyle\lim_{x \to{+}\infty}{\frac{2e^x}{3e^x}}=\frac{2}{3}$$
- $$\displaystyle\lim_{x \to{+}\infty}{\frac{(1-x)\cdot 2^x}{(1+x^2)\cdot 2^x-1}}=\displaystyle\lim_{x \to{+}\infty}{\frac{-x \cdot 2^x}{x^2\cdot 2^x}}=\displaystyle\lim_{x \to{+}\infty}{\frac{-1}{x}}=0$$