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:
$$$\displaystyle\lim_{x \to{+}\infty}{f(x)-g(x)=(\pm \infty) - (\pm \infty)}$$$
and thus, we have an indeterminate form.
To solve for this limit we have three options:
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1.- When one sees the limit straight away: Sometimes it can be seen straight away that one of the two inities is greater than the other:
- $$\displaystyle\lim_{x \to{+}\infty}{x-2^x}=-\infty$$
- $$\displaystyle\lim_{x \to{+}\infty}{\Big(\frac{x-2}{\sqrt{x-1}}- \ln(2x-1)\Big)}=+\infty$$
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2.- When it is possible simplify the expression: Another option is to do the subtraction and to obtain only one expression where probably we will have an indeterminate form of the type infinity divided by infinity, which we can already solve.Let's see an example:
$$$\displaystyle\lim_{x \to{+}\infty}{\Big(\frac{2x^2-5x}{x+3} -2x\Big)}=\displaystyle\lim_{x \to{+}\infty}{\frac{2x^2-5x-2x \cdot (x+3)}{x+3}}=\displaystyle\lim_{x \to{+}\infty}{\frac{-11x}{x+3}}=-11$$$
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3.- When there are square roots involved: If we have $$f(x)$$ or $$g(x)$$ with a square root is not easy to find the limit.
An option is to add expressions in the following way:
$$$f(x)-g(x)=(f(x)-g(x)) \cdot \frac{f(x)^2+g(x)^2}{f(x)+g(x)}$$$
in order to obtain an expression which we can. Let's see an example:
$$$\displaystyle\lim_{x \to{+}\infty}{\sqrt{x^2-2}-x}=\displaystyle\lim_{x \to{+}\infty}{\frac{(\sqrt{x^2-x}-x)\cdot(\sqrt{x^2-x}+x)}{\sqrt{x^2-x}+2}}=$$$
$$$=\displaystyle\lim_{x \to{+}\infty}{\frac{x^2-x-x^2}{\sqrt{x^2-x}+x}}=\displaystyle\lim_{x \to{+}\infty}{\frac{-x}{\sqrt{x^2-x}+x}}=$$$
$$$=\displaystyle\lim_{x \to{+}\infty}{\frac{-x}{x^{\frac{2}{2}}+x}}=\displaystyle\lim_{x \to{+}\infty}{\frac{-x}{2x}}=\frac{-1}{2}$$$