Can $\lim_{x \to 0} \frac{x^3-7x}{x^3}$ be rewritten as $\lim_{x \to 0} \frac{x^3(1-7x^{-2})}{x^3}?$
These two seem to give different answers. Please help me I'm really confused :S
Can $\lim_{x \to 0} \frac{x^3-7x}{x^3}$ be rewritten as $\lim_{x \to 0} \frac{x^3(1-7x^{-2})}{x^3}?$
These two seem to give different answers. Please help me I'm really confused :S
What you did is fine. The next step is to cancel the $x^3$ terms to get
$\lim_{x \to 0} \frac{x^3 - 7x}{x^3} = \lim_{x \to 0} \frac{x^3 (1 - 7x^{-2})}{x^3} = \lim_{x \to 0} 1 - 7x^{-2}$
Now, we have a polynomial over a polynomial and we get division by 0 but the numerator of the fraction is not 0. If it were a left or right hand limit, this would tell us the answer is $+\infty$ or $-\infty$ and we would only need to figure out which one it is. If it's a limit (not right or left hand), there is the possibility that the left and right hand limits differ (one is $+\infty$ and the other is $-\infty$) so the limit could not exist. But, as $x \to 0$, the whole thing $1 - 7x^{-2}$ is going to be negative no matter if you approach from the left or right. Therefore, the left and right hand limits are both $-\infty$, so that the limit itself is $-\infty$.
Your rewrite is correct. The limit of the expression, in either form, does not exist. Alternately, if you allow $\infty$ or $-\infty$ as possible answers to a limit question, the limit is $-\infty$.
A direct way to calculate the limit is to note that if $x\ne 0$, then $\frac{x^3-7x}{x^3}=1-\frac{7}{x^2}.$
They are indeed the same:
$\lim_{x\to 0}\frac{x^3-7x}{x^3}=\lim_{x\to 0}\frac{x^3(1-7x^{-2})}{x^3}=\lim_{x\to 0}\left(1-\frac7{x^2}\right)\;,$ which does not exist. However, it fails to exist in a relatively nice way: as $x\to 0$, the expression in parentheses gets increasingly negative without any bound, so we often write
$\lim_{x\to 0}\left(1-\frac7{x^2}\right)=-\infty\;.$