# The limit $x\to 0$ of a fraction, different answers?

I can solve the following limit: \begin{align} &\lim_{x\to 0}\frac { x(4+x) } { x(x^2+x(C_1+4)+C_2) }\\ &= \lim_{x\to 0}\frac { 4+x } { x^2+x(C_1+4)+C_2 }= \frac{4}{C_2} \end{align} So far so good. But why is the following method wrong (factoring out $x$ in the numerator and denominator)? \begin{align} &\lim_{x\to 0}\frac { x(4+x) } { x(x^2+x(C_1+4)+C_2) }\\ &= \lim_{x\to 0}\frac { x^2(\frac{4}{x}+1) } { x^2(x+C_1+4+\frac{C_2}{x})} \\ &= \lim_{x\to 0}\frac { \frac{4}{x}+1 } { x+C_1+4+\frac{C_2}{x} }\\ &\quad \to \frac{\frac{4}{0}+1}{0+C_1+4+\frac{C_2}{0}}\\ &\quad\to \frac{\infty+1}{0+C_1+4+\infty} \to \frac{\infty}{\infty} \quad \text{indeterminate} \end{align}

• You actually wrote $\frac40\ldots$! – Lord Shark the Unknown Sep 15 '18 at 18:18
• You're right that the result is indeterminate. But, this doesn't mean that you don't get $\frac{4}{C_2}$. You don't stop when you get an indeterminate form. Instead, you have to do some work (such as l'Hopital's rule) to evaluate the limit. – Michael Burr Sep 15 '18 at 18:26
• Apart from the question posed, I think the exercise expects you to find the limits when $C_2\neq 0$ and $C_2=0,C_1\neq-4$ and $C_2=0,C_1=-4$, not just assuming the one case where $C_2$ is non zero. – zwim Sep 15 '18 at 20:06

## 1 Answer

If you invoke the property that limit of product is the product of limits you need both limits to be finite (and exist obviously).

In the expression $$\lim_{x\to 0}\frac { \frac{4}{x}+1 } { x+C_1+4+\frac{C_2}{x} }$$ neither the limit of numerator nor denominator exist. (your wrote $\infty$ , but remember it could also be that $x\to 0-$)