Consider $f(x)=x$ and $g(x)=1/x$. If we take the limit of $f(x)g(x)$ with x tending to $\infty$, why is the limit 1?

At $x=0$, $g(x)$ is undefined, so why should we be allowed to "cancel" out the x in the numerator and denominator? More generally, is $h(x)=f(x)*g(x)=1$? If yes, why?


Note that $$ f(x)g(x)=\begin{cases} 1& \text{if}\quad x\neq0\\ \text{undefined}& \text{if}\quad x=0 \end{cases} $$ In particular, given $\epsilon>0$, if $x>1$, then $|f(x)g(x)-1|=0<\epsilon$ . Thus by the definition of the limit $$ \lim_{x\to\infty} f(x)g(x)=1 $$ Intuitively to take a limit as $x\to \infty$ we only need to know the values of the function above some threshold $M$. What happens below $M$ is irrelevant for the purposes of taking the limit


For values of $x$ where $x\ne0$, $f(x)*g(x)$ will clearly be exactly 1, because for those values there are no problems at all with it being undefined. For example, if $x=5$, then $f(x)*g(x)={5\over 5}$, which is clearly just 1.

As you take the limit to infinity, you can "cancel out the x in the numerator and the demoninator", because for those values that are not zero, the function is very clearly defined. For the special case where $x=0$, you are correct in saying the x's cannot be cancelled out.

$h(x)$ is not defined at $0$ because $g(x)$ is not defined at zero. This, however, does not at all impact our ability to take the limit as $x\to \infty$.

  • $\begingroup$ Thanks. So even if we take the limit as x approaches zero, the cancellation is valid because x is never "equal to" zero in that case? $\endgroup$ – PGupta May 20 '18 at 17:17
  • $\begingroup$ Yes, that's correct. Glad to have helped! $\endgroup$ – volcanrb May 20 '18 at 17:17

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