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**Jhevon** it's kind of hard to explain, you just have to "see" it

for instance, the first. i wanted $\displaystyle \lim_{x \to a}f(x) = \infty$, $\displaystyle \lim_{x \to a}g(x) = 0$

since i want to get $\displaystyle \lim_{x \to a}h(x) = \infty$ i said to myself. how can i get the limit to go to infinty? well, i can "divide by zero" or i can have the function be, say x, or e^x or whatever such that when x goes to infinity, the function goes to infinity, so i chose the latter.

so ok, obviously now i want $\displaystyle \lim_{x \to \infty}$, so i choose $\displaystyle a = \infty$

now i need to get $\displaystyle f(x)$ and $\displaystyle g(x)$

well, i know that $\displaystyle \lim_{x \to \infty} \frac 1x = 0$ because i have seen that limit so many times (and you should have too). so that's my $\displaystyle g(x)$, now, what do i want $\displaystyle f(x)$ to be?

i know $\displaystyle \lim_{x \to \infty}x = \infty$, so could that be $\displaystyle f(x)$? well, no, since i would have $\displaystyle h(x) = x \cdot \frac 1x = 1$ which does not go to infinity as x goes to infinity. so i realize, i just need one more factor of x to get my $\displaystyle \lim_{x \to \infty} x = \infty$, so i made $\displaystyle f(x) = x^2$. so now, $\displaystyle \lim_{x \to \infty}f(x) = \infty$, $\displaystyle \lim_{x \to \infty}g(x) = 0$ and $\displaystyle \lim_{x \to \infty}h(x) = \lim_{x \to \infty}x^2 \cdot \frac 1x = \lim_{x \to \infty}x = \infty$ as desired.

similar reasoning went into the others. just try to think of well-known limits and tweek them to fulfill the conditions. as you see, i did not use any complicated functions (well, the last one was kind of complicated, but only slightly), just keep it simple.