Limit for some general variable I'm having a hard time understanding the solution to this problem.
How is it that, for some general variable '$c$', the limit is always positive infinity, except for the case where $c = 1$?
If $c = 0$, shouldn't the limit be negative infinity?
Further, what about $c = -1$?

 A: Let's first consider when $c=1$:
We have the following:
$$\lim_{t\to0}\frac {5ct+c-1}{5t^2+t}=\lim_{t\to 0}\frac{5t}{5t^2+t}=\lim_{t\to0}\frac{5t}{t(5t+1)}=\lim_{t\to0}\frac{5}{5t+1}=\frac 51=5$$
Note, the limit is not to $\pm\infty$, but instead to a finite value. Thus once we rationalize the original limit, we can simply plug in the value for $t$ and solve.
Let's consider when $c=0$
Then we have:
$$\lim_{t\to0}\frac{5ct+c-1}{5t^2+t}=\lim_{t\to0}\frac{-1}{5t^2+t}=-\infty$$
We cannot rationalize this expression, and thus given t arbitrarily close to $0$, the limit extends to $\infty$. Why? Well we are dividing by an arbitrarily small number!
How about when $c=-1$
Then we see:
$$\lim_{t\to0}\frac{5ct+c-1}{5t^2+t}=\lim_{t\to0}\frac{-5t-2}{5t^2+t}=-\infty$$
Again, this expression cannot be rationalized, and thus we must consider what happens for $t$ arbitrarily close to $0$. The numerator approaches $-2$, while the denominator approaches infinitely to $0$.
Now we can generalize this for all $c\in \mathbb{R}$:
For $c>1$, $\lim_{t\to0}\frac{5ct+c-1}{5t^2+t}=\infty$, for $c<1$, $\lim_{t\to0}\frac{5ct+c-1}{5t^2+t}=-\infty$, and for $c=1$, $\lim_{t\to0}\frac{5ct+c-1}{5t^2+t}=5$.
A: There is a problem in dealing with $c\neq 1$. In this case the denominator tends to $0$ but does not maintain a constant sign. The numerator tends to a nonzero constant $c-1$. So in this case the function oscillates infinitely. However if we deal with one sided limits say when $t\to 0^{+}$ then the function tends to $\infty$ ($-\infty$) if $c>1$ (if $c<1$). The case $t\to 0^{-}$ can be dealt in a similar manner. 
