# Thread: I don't understand why the limit is negative infinity

1. ## I don't understand why the limit is negative infinity

Hello everyone,

I tried to solve it and in my solution, I get positive infinity when I approach from the left (which is correct, according to the answer) and positive infinity which is false when I approach from the right.

That is the solution:

Can someone explain, how I get the results?

2. ## Re: I don't understand why the limit is negative infinity

Originally Posted by ORT

I tried to solve it and in my solution, I get positive infinity when I approach from the left (which is correct, according to the answer) and positive infinity which is false when I approach from the right.
That is the solution: Can someone explain, how I get the results?
The numerator $x^3-3x^2 \approx -4$ for each $x\approx 2$ regardless of right or left.
But the denominator $x-2<0$ for $x<2$ while $x-2>0$ for $x>2$

3. ## Re: I don't understand why the limit is negative infinity

Thank you very much!

4. ## Re: I don't understand why the limit is negative infinity

Originally Posted by ORT
Hello everyone,

I tried to solve it and in my solution, I get positive infinity when I approach from the left (which is correct, according to the answer) and positive infinity which is false when I approach from the right.

That is the solution:

Can someone explain, how I get the results?
\displaystyle \begin{align*} \lim_{x \to 2} \frac{x^3 - 3\,x^2}{x - 2} &= \lim_{x \to 2} \left( \frac{x^3 - 2\,x^2 - x^2}{ x - 2 } \right) \\ &= \lim_{x \to 2} \left( \frac{x^3 - 2\,x^2}{x - 2} - \frac{x^2}{x - 2} \right) \\ &= \lim_{x \to 2} \left[ \frac{x^2 \left( x - 2 \right) }{x - 2} - \frac{x^2}{x- 2} \right] \\ &= \lim_{x \to 2} \left( x^2 - \frac{x^2}{x - 2} \right) \\ &= \lim_{x \to 2} \left( x^2 + \frac{-x^2 + 2\,x - 2\,x}{x - 2} \right) \\ &= \lim_{x \to 2} \left[ x^2 + \frac{-x\left( x - 2 \right) - 2\,x}{x - 2} \right] \\ &= \lim_{x \to 2} \left[ x^2 - \frac{x \left( x - 2 \right) }{x - 2 } - \frac{2\,x}{x - 2} \right] \\ &= \lim_{x \to 2} \left( x^2 - x - \frac{2\,x}{x - 2} \right) \\ &= \lim_{x \to 2} \left( x^2 - x + \frac{-2\,x + 4 - 4}{x - 2} \right) \\ &= \lim_{x \to 2} \left[ x^2 - x + \frac{-2\,x + 4}{x - 2} - \frac{4}{x - 2} \right] \\ &= \lim_{x \to 2} \left[ x^2 - x + \frac{-2 \left( x - 2 \right) }{x - 2} - \frac{4}{x - 2} \right] \\ &= \lim_{x \to 2} \left( x^2 - x - 2 - \frac{4}{x - 2} \right) \end{align*}

The \displaystyle \begin{align*} x^2 - x - 2 \to 0 \end{align*} as \displaystyle \begin{align*} x \to 2 \end{align*} so the only important term is \displaystyle \begin{align*} -\frac{4}{x - 2} \end{align*}. As \displaystyle \begin{align*} x \to 2^- \end{align*} we have \displaystyle \begin{align*} -\frac{4}{x - 2} \to +\infty \end{align*} and as \displaystyle \begin{align*} x \to 2^+ \end{align*} we have \displaystyle \begin{align*} -\frac{4}{x - 2} \to -\infty \end{align*}.

Thus the limit does not exist.