Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Show that if $\lim_{x\rightarrow x_0} f(x)=L>0$ then $\lim_{x\rightarrow x_0} 1/f(x)=1/L>0$

My proof is right now using the fact that $\delta = \epsilon/L$. I'm not sure whether this is correct at all.

share|cite|improve this question
up vote 1 down vote accepted

Given $\epsilon>0$ you want to find $\delta>0$ such that $|x-x_0|<\delta$ implies $\left|\frac1{f(x)}-\frac1L\right|<\epsilon$. Observer that $\frac1{f(x)}-\frac1L = \frac{f(x)-L}{f(x)L}$. You can make sure that the numerator is small because $f(x)\to L$. But how can you prevent the denominator from getting small (which migh tmake the fraction big)? By a suitable choice of $\delta$ you can make sure that $f(x)>\frac L2$ (namely, that $|f(x)-L|<\frac L2$). This makes $\left|\frac1{f(x)}-\frac1L\right|<\frac2{L^2}\cdot |f(x)-L|$, hence "manageable". You must make your $delta$ small enough to fulfill two conditions, that is let $\delta=\min\{\delta_1,\delta_2\}$ where $\delta_1$ is chosen to warrant $f(x)>\frac L2$ and $\delta_2$ is chosen to warrant $|f(x)-L|<\frac{L^2}2\cdot\epsilon$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.