0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $|f(x)-L|<\frac{L^2}2\cdot\epsilon$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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