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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 $c = -1$?

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Widawensen
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darylnak
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  • The limit at $0$ of a rational function is the limit of the ratio of the lowest degree terms of it numerator and denominator. – Bernard Apr 23 '17 at 10:30
  • When $c\ne 1$, the numerator goes to $c-1 \ne 0$ while the denominator goes to 0, resulting in the fraction exploding to infinity. Here the point is not that it is $+\infty$, but tht it diverges. The infinity to which the limit is depends on wether you are approaching 0 with positive or negative values here – Astyx Apr 23 '17 at 10:30

2 Answers2

1

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 $c=1$, $\lim_{t\to0}\frac{5ct+c-1}{5t^2+t}=5$.

Mark Pineau
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0

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.

Paramanand Singh
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