Suppose $a$ and $b$ are two real numbers with $a<0$ and $b>0$, and $f$ is a continuous function with $\lim_{x\to +\infty} f(x)= b$ and $\lim_{x\to -\infty} f(x)= a$. How can I prove that the equation $f(x)=0$ has at least one solution in $\mathbb{R}$ using the definition of the limit?
Existence of zero using definition of limit
2
$\begingroup$
real-analysis
limits
1 Answers
3
If $\lim_{x\to\infty}f(x)=b>0$, then there is some $N\in R$ such that $x>N$ implies $\frac12 b=b-\frac12 b
If $\lim_{x\to-\infty}f(x)=a<0$, then there is some $M\in R$ such that $x
Now $f$ is continuous and $f(p)<0 and $f(q)=0$.
-
0thanks, now it's clear and good – 2012-11-05