Continuity with limits

**Ife** · April 16th 2010, 06:15 AM

I am having some trouble with the concept of continuity and limits when it applies to questions. Here's a simple question I am not sure how to approach:

$\mbox{If} x^4 \leq \mbox{f(x)} \leq x^2$ for $-1 \leq x \leq 1$ and $x^2 \leq \mbox{f(x)} \leq x^4$ for $x<-1 \mbox{or} x>1$ , at what points $\mbox{c}$ do you automatically know $\lim_{x\to\mbox{c}}\mbox{f(x)}$ ?

What are the values of the limits at these points?

What is the correct approach to this question?

**Diego** · April 16th 2010, 01:35 PM

In x = 0, -1, 1 . You can tell what the limits are.

In x = 0, the limit is zero.
In x = 1, -1 the limit is one.

It is a consequence of the sandwich theorem:

$\lim_{x \to 0} x^{4} = \lim_{x \to 0} f(x) = \lim_{x\to 0} x^{2}$

the same can be said in x = -1,1 though in these cases you have to consider lateral limits.

**Ife** · April 17th 2010, 02:58 AM

Originally Posted by Diego

In x = 0, -1, 1 . You can tell what the limits are.

In x = 0, the limit is zero.
In x = 1, -1 the limit is one.

It is a consequence of the sandwich theorem:

$\lim_{x \to 0} x^{4} = \lim_{x \to 0} f(x) = \lim_{x\to 0} x^{2}$

the same can be said in x = -1,1 though in these cases you have to consider lateral limits.

Thanks.. though i am not sure I know about Lateral Limits...

**HallsofIvy** · April 17th 2010, 04:57 AM

I believe he means "one-sided" (from the left and from the right) limits.

**Diego** · April 17th 2010, 06:45 AM

Oh, sorry I meant one-sided limits (in spanish they are called lateral). For instance in x=1, if you approach the limit with numbers bigger than one (where $x^{2} < f(x) < x^{4}$ ):

$\lim_{x \to 1^{+}} x^{2} = \lim_{x \to 1^{+}} f(x) = \lim_{x\to 1^{+}} x^{4}=1$

and if smaller than one (where $x^{4} < f(x) < x^{2}$ , if x is not too far from 1):

$\lim_{x \to 1^{-}} x^{4} = \lim_{x \to 1^{-}} f(x) = \lim_{x\to 1^{-}} x^{2} =1$

thus:

$\lim_{x \to 1} f(x) = 1$

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Continuity with limits

Use sandwich theorem.

Lateral limits

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