Computing $\lim_{x\to (3/2)^-} (x^2+1)/(3x-2x^2)$

Question

I am trying to compute $$\large \lim_{x\to \frac32^-}\frac{x^2+1}{3x-2x^2}$$

I know the numerator will equal one fine and how it becomes $x^2+1 * 1/(3x-2x^2)$ because $a/b = a * 1/b$, but then I'm not sure where I go from there, I know the answer is positive infinity but once I get $1/3x-2x^2$ I try to plug $3/2$ in and that gets $0$, I then plugin slightly over or under $3/2$ like $1.6$ or $1.4$ and I get valid outputs. I had the same problem earlier except it approached 0 from the left side and it was negative infinity, I have absolutely no idea what to do or where to go from this.

Answer 1

HINT

Note that

$$3x-2x^2=2x\left(\frac32-x\right)$$

Answer 2

The numerator will not equal $1$, it will approach $1+(\frac 32)^2=\frac {13}{4}$. The denominator will approach $0$ but always be positive because $x \lt \frac 32$. The ratio will therefore go to $+\infty$. A graph is below. The fact that $x$ approaches from below means you are on the lower piece of the graph, which goes to $+\infty$

Answer 3

For such a problem, let $x=y+\frac 32$ to make $$ \lim_{x\to \frac32}\frac{x^2+1}{3x-2x^2}=-\lim_{y\to0}\frac{13+12y+4 y^2}{12y+8 y^2}=-\lim_{y\to0}\frac{13+12y+4 y^2}{12y(1+\frac 23 y)}$$ Since $y$ is small, the numerator is almost $13$ and the denominator almost $12y$.

Answer 4

Graphical way to observe the denominator $3x-2x^2$:

$\hspace{3cm}$

It implies that as $x\to 1.5$, $3x-2x^2\to 0+$.

Hence: $$\lim_{x\to \frac32-}\frac{x^2+1}{3x-2x^2}=\frac{13/4}{0+}=+\infty.$$