Take the 2-minute tour ×
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.

Any help in solving the following problem would be greatly appreciated:

Let $f, g_1, g_2$ be functions from $\mathbb R$ to $\mathbb R$, with $g_1(x) \leq f(x) \leq g_2(x)$, for all $x \in \mathbb R$. Suppose that, for some $p \in \mathbb R$, we have $\lim_{x \rightarrow p} g_1(x) = \lim_{x \rightarrow p} g_2(x) = c$. Show that $ \lim_{x\rightarrow p} f(x) = c,$ as well.

I've spent hours on it and haven't come up with anything worth posting.

share|improve this question
1  
What do you mean by $\lim_{x \rightarrow p} > g_1(x) $? –  sidht Nov 30 '12 at 19:57
    
$\lim_{x \rightarrow p} > c$ ? –  Charlie Nov 30 '12 at 20:02
    
@sizz: I suspect that the $>$ is just a typo. –  Brian M. Scott Nov 30 '12 at 20:03
2  
isn't it the "squeeze theorem" for limits?if $g_1$ and $g_2$ have limit equal $c$ then $f$ has limit equal $c$? –  Charlie Nov 30 '12 at 20:05
    
@Charlie, in that case web.mit.edu/wwmath/calculus/limits/squeeze.html –  sidht Nov 30 '12 at 20:08
show 2 more comments

1 Answer

up vote 4 down vote accepted

HINT: Let $\epsilon>0$. There are real numbers $\delta_1,\delta_2>0$ such that $$|g_1(x)-c|<\frac{\epsilon}2\quad\text{whenever}\quad 0<|x-p|<\delta_1$$ and $$|g_2(x)-c|<\frac{\epsilon}2\quad\text{whenever}\quad 0<|x-p|<\delta_2\;.$$

Now show that

$$|f(x)-c|<\epsilon\quad\text{whenever}\quad 0<|x-p|<\min\{\delta_1,\delta_2\}\;.$$

share|improve this answer
    
@BrainMScott Thank you! The procedure seems simple when one examines it. –  Neil Dec 3 '12 at 18:29
    
@Neil: You’re welcome. Experience helps in seeing where to start, but I agree that the basic idea isn’t hard. –  Brian M. Scott Dec 3 '12 at 21:07
add comment

Your Answer

 
discard

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.