If \(\lim_{x\rightarrow 0}\frac{f(x)}{{x}^{2}}=5\), find the following limits.

(a) \(\lim_{x\rightarrow 0} f(x)\) (b) \(\lim_{x\rightarrow 0}\frac{f(x)}{x}\)

How do I solve this problem from my Calculus I and II college math book. Please show how you get to the answer step by step.

gibsonj338
Sep 28, 2018

#1**+1 **

\(\text{suppose }f(x) \text{ is a well behaved function in that it has a Taylor series}\\ f(x) = \sum \limits_{k=0}^\infty~c_k \dfrac{x^k}{k!}\)

\(\dfrac{f(x)}{x^2} = \sum \limits_{k=0}^\infty~c_k\dfrac{x^{k-2}}{k!}=\dfrac{c_0}{x^2}+\dfrac{c_1}{x}+\dfrac{c_2}{2}+\dots \text{ function of powers of x}\)

\(\text{In order for }\lim \limits_{x\to 0}\dfrac{f(x)}{x^2} = 5 \\ c_0=0,~c_1=0~c_2 = 10 \\ \text{so } f(x) = 5x^2+\text{higher powers of x}\)

a) \(\lim \limits_{x \to 0} f(x) = c_0 = 0\)

b) \(\lim \limits_{x\to 0}\dfrac{f(x)}{x} = c_1 = 0\)

Rom
Sep 28, 2018

#2**0 **

Not all functions can be described by a taylor series.

Besides, this question doesn't require taylor series at all.

Guest Sep 29, 2018

#5**0 **

What Melody means is demonstrate your method without using the Taylor series or shut the fuckup!

Guest Sep 29, 2018

#6**+1 **

Why should I? Rom had a wrong assumption and because of that assumption he gave a partial solution (although the final answers were correct). I'm not blaming or mocking him for that beacuse I could've easily made that assumption too, But it's also worth noting that this level of math is not required for this question. Sometimes, an easy solution may not be that obvious, but in this case i think using a taylor series is just an overcomplexification of the question. A student asking this question might not know/remember what taylor sums are, and it makes the question seem harder than it is.

So I can answer the question now and I can break my own fingers too, but I don't like spoiling solutions and I would rather let rom try this question now that he knows that his assumption is wrong.

But because you're all so eager, I'll give my solution-

\(\lim_{x\rightarrow 0} \frac{f(x)}{x^2}=5\Rightarrow \lim_{x\rightarrow 0} \frac{f(x)}{x^2}-5=0 \\ \lim_{x\rightarrow 0} x^2=0 \Rightarrow \lim_{x\rightarrow 0} (\frac{f(x)}{x^2}-5)*(x^2)= \\(\lim_{x\rightarrow 0} \frac{f(x)}{x^2}-5) *(\lim_{x\rightarrow 0} x^2)=\\ 0*0=0\\\)

\(0=\lim_{x\rightarrow 0} (\frac{f(x)}{x^2}-5)*(x^2)=\lim_{x\rightarrow 0} f(x)-5x^2\Rightarrow\\ 0=0+\lim_{x\rightarrow 0} 5x^2=\lim_{x\rightarrow 0} f(x)-5x^2+\lim_{x\rightarrow 0} 5x^2=\lim_{x\rightarrow 0} f(x)\)

The solution to section b is almost identical.

Guest Sep 29, 2018

#7**+1 **

*Rom had a wrong assumption and because of that assumption he gave a partial solution (although the final answers were correct).*

You are full of Bullshit!

Rom made an assumption and it was correct. He gave a partial solution because he wants the student to learn from this and not just parrot answers. Now the student knows more about how and when to apply the Taylor series. Applying a complex process to simple questions is an optimal way to learn the rote mechanics.

The Taylor series isn’t a “complexification,” it’s actually a simplification. The Taylor series gives an analytical understanding to paradoxical mathematics such as those found in Zeno's paradoxes. This makes complex problems less complex, not more complex.

GA

GingerAle
Sep 29, 2018

#8**0 **

I understand that rom may have done that because he wanted gibson to understand what the answer is without giving out the method, sometimes having extra assumptions just for understanding what the final answer is supposed to be can be helpful, and I understand that analytic functions are very nice to work with, but again, not all functions are analytic and this solution doesn't apply. Choosing f(x)=5*x^{2} is enough for showing that f is supposed to converge to 0 at 0.

This level of math is not needed for this question, and it doesn't solve this question.

Guest Sep 30, 2018

edited by
Guest
Sep 30, 2018