Prove the claim in Limits using definitions. If $\lim_{x\to c}f(x) \neq 0$ and $ \lim_{x\to c} g(x)=0$, show that $\lim_{x\to c}\frac{f(x)}{g(x)}$ doesnot exist as a finite value.
I have wrote the definitions of limit for the limits of $f(x)$ and $g(x)$. Then I assumed that  $\lim_{x\to c}\frac{f(x)}{g(x)}$ exist as a finite value and tried get a contradiction. But I was unable to find the solution.
Please help.
 A: It is much simpler to use algebra of limits. Let us suppose on the contrary that $f(x)/g(x)$ does tend to a finite limit (say $L$) as $x \to c$. Then we know that $$\lim_{x \to c}f(x) = \lim_{x \to 0}\frac{f(x)}{g(x)}\cdot g(x) = L \cdot 0 = 0$$ and this contradicts our hypothesis. It follows that $f(x)/g(x)$ can not tend to a finite limit.
If we don't want to use the algebra of limits and instead use the $\epsilon-\delta$ definition of limit we can proceed as follows. Let us assume (like before) that $f(x)/g(x)$ does tend to a finite limit (say $L$) as $x \to c$. Hence we have a $\delta_{1} > 0$ such that $$\left|\frac{f(x)}{g(x)} - L\right| < 1$$ whenever $0 < |x - c| < \delta_{1}$. Therefore $$\left|\frac{f(x)}{g(x)}\right| < |L| + 1 = A\text{ (say)}$$ Let $\epsilon > 0$ be an arbitrary number. Since $g(x) \to 0$ as $x \to c$ it follows that there is a number $\delta_{2} > 0$ such that $$|g(x)| <\frac{\epsilon}{A}$$ whenever $0 < |x - a| < \delta_{2}$.
Let $\delta = \min(\delta_{1}, \delta_{2})$. Then for $0 < |x - c| < \delta$ we have $$|f(x)| = \left|\frac{f(x)}{g(x)}\cdot g(x)\right| < A\cdot\frac{\epsilon}{A} = \epsilon$$ and hence it follows that $f(x) \to 0$ as $x \to c$ and this contradicts our hypothesis. It follows that $f(x)/g(x)$ can not tend to a finite limit as $x \to c$.
A: From $\lim_{x\to c} f(x) \neq 0$ we know that:
$$\exists \epsilon_0 > 0,\,\,\, \forall \delta > 0,\,\,\, \exists x \text{ with } 0 < \lvert x-c\rvert < \delta \,\, \text{ and } \,\, \lvert f(x) \rvert > \epsilon_0$$
From $\lim_{x\to c}g(x)=0$ we know that
$$\forall \epsilon > 0,\,\,\, \exists \delta > 0,\,\,\, \forall x \text{ with } 0 < \lvert x-c\rvert < \delta, \,\,\, \lvert g(x) \rvert < \epsilon$$
If for any punctured neighborhood $N$ of $c$ there is some $x \in N$ with $g(x)=0$, then the limit does not exist (because the quotient is not defined). In this case, there is nothing to prove; let us then assume $g\neq 0$ near $c$. With this in mind, we will use the equivalent inequality $\left\lvert\frac{1}{g(x)}\right\rvert > \frac{1}{\epsilon}$.
Now, suppose there were a finite limit $L$ for $\frac{f(x)}{g(x)}$ as $x \to c$. We will assume for the sake of simplicity that $L>0$, but the argument can be easily adapted to the case $L<0$. Proceeding, if this limit were true we'd have that:
$$\forall \epsilon > 0,\,\,\, \exists \delta > 0,\,\,\, \forall x \text{ with } 0 < \lvert x-c\rvert < \delta, \,\,\, \left\lvert \frac{f(x)}{g(x)}-L \right\rvert < \epsilon$$
Equivalently:
$$\forall \epsilon > 0,\,\,\, \exists \delta > 0,\,\,\, \forall x \text{ with } 0 < \lvert x-c\rvert < \delta, \,\,\, L-\epsilon < \frac{f(x)}{g(x)}  < L+\epsilon$$
We show the statement above does not hold, so that our assumption of existence is false.
Let $k>0$ be such that $L+\frac{\epsilon_0}{kL} < kL$; it is clear that such a $k$ exists (what happens as $k$ becomes very large?). Now, choose $\epsilon$ with $0 < \epsilon < \frac{\epsilon_0}{kL}$. If our fourth statement were true, there would be some $\tilde{\delta}>0$ such that any $x$ with $0 < \lvert x-c\rvert < \tilde{\delta}$ satisfied $\frac{f(x)}{g(x)}  < L+\epsilon$.
Our second statement guaratees that there is some $\delta_0>0$ with the property that any $x$ with $0 < \lvert x-c\rvert < \delta_0$ satisfies $\left\lvert\frac{1}{g(x)}\right\rvert > \frac{1}{\epsilon} > \frac{kL}{\epsilon_0}$. However, our first statement says there is some $x_0$ with $0 < \lvert x_0-c\rvert < \min\{\tilde{\delta},\delta_0\}$ and $|f(x_0)|>\epsilon_0$. Hence: $$\left\lvert\frac{f(x_0)}{g(x_0)}\right\rvert > kL>L+\frac{\epsilon_0}{kL}>L+\epsilon,$$
contradicting that all $x$ with $0 < \lvert x-c\rvert < \tilde{\delta}$ satisfies $\frac{f(x)}{g(x)}  < L+\epsilon$.
I know the text may have sounded hyper-redundant or overly complicated, but I tried to be as clear as and basic as possible as this seems like an introductory analysis exercise. It's really a matter of juggling epsilons and deltas with the definitions (and their negations).
