1. Left hand limit

Since I haven't learned to use LaTex yet, for convenience, let L.H.f(x) be the limit of f as x approaches c from the left.

1.) If f(x)>0 and L.H.f(x) = 0, show that L.H.(1/f(x)) = infinity.
2.) Give an example of a function f for which L.H.f(x) = 0, but L.H.(1/f(x)) is neither infinity nor negative infinity.

2. Originally Posted by spoon737
Since I haven't learned to use LaTex yet, for convenience, let L.H.f(x) be the limit of f as x approaches c from the left.

1.) If f(x)>0 and L.H.f(x) = 0, show that L.H.(1/f(x)) = infinity.
2.) Give an example of a function f for which L.H.f(x) = 0, but L.H.(1/f(x)) is neither infinity nor negative infinity.
Hints: function is positive, but it goes to zero as x goes from negative infinity to c, so it must be a decreasing function in that domain. Therefore, 1/f(x) will go to infinity. To show this, all you need to do is use the basic properties of limits.

$\displaystyle lim_{ x\rightarrow a} \frac{f(x)}{g(x)} = \frac{lim_{ x\rightarrow a} f(x)}{lim_{x\rightarrow a} g(x)}$

In this case (#1):

$\displaystyle lim_{ x\rightarrow a} \frac{1}{f(x)} = \frac{lim_{ x\rightarrow a} 1}{lim _{x\rightarrow a} f(x)} \Rightarrow \frac{1}{0} = \infty$

3. Thanks for your quick response. I was thinking the proof would be similar to yours, but I wasn't sure if that reasoning alone would be rigorous enough (this is for an analysis class). I was thinking I needed to use the epsilon-delta definition in my proof.

4. Originally Posted by spoon737
Since I haven't learned to use LaTex yet, for convenience, let L.H.f(x) be the limit of f as x approaches c from the left.

1.) If f(x)>0 and L.H.f(x) = 0, show that L.H.(1/f(x)) = infinity.
If $\displaystyle \lim_{x\to c} f(x) = \infty$ it means for any $\displaystyle N>0$ there exists a $\displaystyle X> 0$ such that if $\displaystyle x>X$ then $\displaystyle f(x) > N$. This means that $\displaystyle 0< 1/f(x) < 1/N$. Thus, $\displaystyle |1/f(x)| < 1/N$ for $\displaystyle x>X$. Let $\displaystyle \epsilon > 0$ then we can choose $\displaystyle N$ so large that $\displaystyle 1/N < \epsilon$ and so $\displaystyle |1/f(x)| < \epsilon$. Thus, if $\displaystyle \lim_{x\to c}f(x) = \infty$ for $\displaystyle f(x) > 0$ it means $\displaystyle \lim_{x\to c}1/f(x) = 0$. Can you imitate this proof to prove what you are trying to show?

5. Originally Posted by ThePerfectHacker
If $\displaystyle \lim_{x\to c} f(x) = \infty$ it means for any $\displaystyle N>0$ there exists a $\displaystyle X> 0$ such that if $\displaystyle x>X$ then $\displaystyle f(x) > N$.
That is for when f(x) tends to infinity as x itself tends to infinity. This question is slightly different: we say that f(x) tends to infinity as x tends to c from the left iff for any N > 0, there exists a δ > 0 such that if c−δ < x < c then f(x) > N.

6. Given $\displaystyle f(x) = \left\{ {\begin{array}{rl} { - x} & {x \in Q^ - } \\ x & {x \in \Re ^ - \backslash Q} \\ \end{array}} \right.$
Then what are these $\displaystyle \lim _{x \to 0^ - } f(x)\,\& \,\lim _{x \to 0^ - } \frac{1}{{f(x)}}$ ?

7. Originally Posted by JaneBennet
That is for when f(x) tends to infinity as x itself tends to infinity. This question is slightly different: we say that f(x) tends to infinity as x tends to c from the left iff for any N > 0, there exists a δ > 0 such that if c−δ < x < c then f(x) > N.
I know. I was doing a different problem because I did not realize he asked that.