$\displaystyle \lim_{ x \rightarrow -7^{-}} \; \frac{1}{x^{2}(x+7)}$
My manual computation yields a different answer to that of Wolfram's.
And I don't understand, why isn't the correct answer: the function approaches $0$? Because that's what it does!
If denominator is expanded, then that's how the whole rational equation looks like:
$\frac {1}{x^3 + 7x^2}$
Now, since $x$ approaches $-7$ from the left, let's take a very small values that is close to $-7$ from the left. Let it be $-7.000000000001$.
$\frac {1}{-7.000000000001^3 + 7 \cdot -7.000000000001^2} = \frac {1}{-686.000000000245000000000028000000000001}$
$1$ divided by a big number? A very SMALL number, meaning the function is approaching $0$! No? Or the function approaches a very small number, whatevs.
Why does WolframAlpha says that the function approaches infinity?
The only possible answer is that I'm not supposed to expand denominator, but rather keep it as factor-y as possible. Then yeah, I guess, in denominator, $x^2$ yields a big number multiplied by a very small number $(x+7)$, then $1$ divided by a very small number does indeed "shoots off" to infinity.
You forgot the parentheses in your calculation. It should be: $$\frac 1{\color{green}{(}-7.000000000001\color{green}{)}^3+7\cdot\color{red}{(}-7.000000000001\color{red})^2}$$ Notice that I put one pair of parentheses in $\color{green}{\text{green}}$ for your cubic power and the other pair of parentheses in $\color{red}{\text{red}}$ for your square power.
Cubing a negative number does not remove the negative sign inside the parentheses, but squaring does. In general, raising a negative number to an odd power remove the negative sign inside the parentheses, but raising a negative number to an even power does.