Specifically, if $\lim\limits_{x \to a} g(x) = b$, and if $f$ is continuous at$ b$, then the $\lim\limits_{x \to a} f \circ g(x) = f(b)$.
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
We need to show given $\epsilon > 0$, there exists a $\delta > 0$ so that $|x-a| < \delta \Rightarrow |f(g(x))-f(b)|<\epsilon$. So first find $\delta' > 0$ so that $|z-b| < \delta' \Rightarrow |f(z)-f(b)| < \epsilon$ (we can find a $\delta'$ because $f$ is continuous at $b$). Then find $\delta > 0$ so that $|x-a| < \delta \Rightarrow |g(x)-g(a)| < \delta'$ (we can find a $\delta$, because $g$ is continuous at $a$). Now collecting this together, if $|x-a| < \delta$ then $|g(x)-g(a)|=|g(x)-b| < \delta'$. But ("taking" $z=g(x)$), this means that $|f(g(x))-f(b)| < \epsilon$, as desired. |
|||
|
|
