Proving a function has a limit at 0, and finding that limit I have the function $ f:(0,1) \to R$ by $f(x)={x^3+6x^2+x\over x^2-6x}$.
I guess the limit to be -1/6 at 0.
I'm having trouble finding $\delta$ for all $0<\lvert x-0\rvert < \delta $ such that $\lvert f(x)+{1 \over 6}\rvert<\epsilon$.
May someone point me in the right direction in finding delta? Is it obvious and I'm just missing it?
So far, simplifying $ \lvert f(x)-L\rvert < \epsilon $ I have $\lvert {x^2+6x+1\over x-6}+{1\over 6}\rvert =\lvert{x(6x+37)\over 6(x-6)}\rvert<\epsilon$. Thank you.
Edit: Yes, I'm trying to prove using the $\epsilon-\delta$ definition. Thanks.
Edit 2: Thank you for the input guys/gals, really cleared it up.

Notice that since the domain of $f$ is $(0,1)$ (i.e., $0 < x < 1$), then $\left |  \dfrac{x(6x + 37)}{6(x - 6)}  \right | \leq  \left |  \dfrac{1(6x + 37)}{6(0 - 6)}  \right | = \left | \dfrac{6x + 37}{36} \right | \leq \frac{1}{6} |x| + \frac{37}{36}$, right?
And we want to find $\delta > 0$ such that $|x| < \delta$ implies the left hand side is $< \epsilon$.
Well, let $\delta = \min \{ 1, 6(\epsilon - \frac{37}{36}) \}$.  Then since as we just showed above:
$\left |  \dfrac{x(6x + 37)}{6(x - 6)}  \right | \leq \frac{1}{6} |x| + \frac{37}{36}$
if $|x| < 6(\epsilon - \frac{37}{36})$, then we get:
$\left |  \dfrac{x(6x + 37)}{6(x - 6)}  \right | \leq \frac{1}{6} |x| + \frac{37}{36} < \frac{1}{6} (6(\epsilon - \frac{37}{36})) + \frac{37}{36} = \epsilon$
and this is what you wanted.
