-1
$\begingroup$

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}

$\endgroup$
  • 8
    $\begingroup$ You actually wrote $\frac40\ldots$! $\endgroup$ – Lord Shark the Unknown Sep 15 '18 at 18:18
  • $\begingroup$ 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. $\endgroup$ – Michael Burr Sep 15 '18 at 18:26
  • $\begingroup$ 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. $\endgroup$ – zwim Sep 15 '18 at 20:06
1
$\begingroup$

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 $x\to 0-$)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.