Why can't this limit be taken before simplification?

Question

Assume $$f(x)=1-x^2+{x^4\over2!}-{x^6\over3!}+{x^8\over4!}+\cdots+{(-1)^nx^{2n}\over n!}.$$

I am trying to find $$\lim_{x\to 0}{1-x^2-f(x) \over x^4}.$$

I first found $\lim_{x\to0}{f(x)}$, which is clearly just $1$, seen after plugging in $0$ for all $x$ in the expression for $f(x)$.

This leaves $\lim_{x\to0}{1-x^2-1 \over x^4}$, which simplifies to $\lim_{x\to0}{1 \over x^2}$. This limit is equal to $-\infty$, however it is not the correct answer.

To find the correct answer, one must first substitute in $f(x)$ and simplify before taking the limit.

This gives $\lim_{x\to0}{1-x^2-(1-x^2+{x^4\over2!}-{x^6\over3!}+{x^8\over4!}+\cdots+{(-1)^nx^{2n}\over n!})\over x^4}$, which can then be simplified to $\lim_{x\to0}{-{x^4\over2!}+{x^6\over3!}-{x^8\over4!} \over x^4}$, and finally $\lim_{x\to0}({{-1\over2}+{x^2\over3!}-{x^4\over4!}+\cdots})$. Plugging in zero leaves only $-{1\over2}$, which is the correct answer.

Why does the second method work while the first method doesn't? Can't limits be distributed around to all the individual $x$s? Is this due to the nature of the infinite series?

Note: This problem comes from the 2007 AP BC Calculus Exam, Free Response Question #6. This question can be found at this link.

Any help would be greatly appreciated!

This might be just a more confusing way of thinking about it, but that function looks like a Taylor's series expansion from somewhere. Maybe you should try and think of it that way somehow... I'm not sure exactly how though. — theREALyumdub, May 02 '18 at 02:11
I believe it is actually the expansion for $e^{-x^2}$, and if I plug in 0 for x, then it is still equal to 1, which keeps the first method the same. — volcanrb, May 02 '18 at 02:16
You can only split limits termwise when you know all the pieces exist. — Randall, May 02 '18 at 02:19
The mistake was in not seeing that x^4/2!/x^4 evaluates to an actual number before taking the limit. — Phil H, May 02 '18 at 02:31

score 2 · Accepted Answer · answered May 02 '18 at 02:21

This is not correct, for the same reason as the thing below:

Since $\lim_{x\to 0} x = 0$, $$ \lim_{x\to 0} \frac{x}{x} = \lim_{x\to 0}\frac{0}{x} =0 $$

is false.

Do you notice the problem? Operations on limits tell use that, if $\lim_{x\to 0} f(x)$, $\lim_{x\to 0} g(x)$, and $\lim_{x\to 0} \frac{f(x)}{g(x)}$ all exist and $\lim_{x\to 0} g(x) \neq 0$, then $$ \lim_{x\to 0} \frac{f(x)}{g(x)} = \frac{\lim_{x\to 0} f(x)}{\lim_{x\to 0} g(x)}\,. \tag{1} $$ Similarly, if $\lim_{x\to 0} f(x)$ and $\lim_{x\to 0} g(x)$ both exist, then $$ \lim_{x\to 0} f(x)+ g(x)= \lim_{x\to 0} f(x)+ \lim_{x\to 0} g(x) \tag{2} $$

But absolutely nothing ever said that $$ \lim_{x\to 0} \frac{f(x)+g(x)}{h(x)} = \lim_{x\to 0} \frac{f(x)+\lim_{y\to 0}g(y)}{h(x)} $$ ! That's not a result that you can invoke nor use -- it's simply not a result.

Thanks. I think the specific part that I was missing was the exception when $\lim_{x \to 0}{g(x)}=0$. Since $x^4$ in the denominator does approach 0, the limit can't be distributed. — volcanrb, May 02 '18 at 02:24

score 0 · Answer 2 · answered May 02 '18 at 02:29

0

You are splitting your function, $${1-x^2-f(x) \over x^4}={1-f(x) \over x^4} -{x^2\over x^4}$$ and none of the two fractions on the right side converges.

answered May 02 '18 at 02:29

Mohammad Riazi-Kermani

68,728

Why can't this limit be taken before simplification?

2 Answers2