4chan archive /sci/ (index)
2012-09-06 01:47 5030961 Anonymous (526157_454544677918664_1777557823_n (1).jpg 500x649 39kB)
hey /sci/ I though this would be oscillating, therefore the limit does not exist, but apparently I'm wrong. Can someone explain this to me?

0 min later 5030966 Anonymous (Untitled.png 1533x714 39kB)
fuck

4 min later 5030980 Anonymous
For extremely large values of x, think x = a billion, 1/x is super close to 0. So while cos(x) is oscillating, 1/x gets smaller and smaller and smaller. cos(0) = 1

8 min later 5030991 Anonymous (Untitled.png 1533x714 31kB)
>>5030980 Ahhh thank you, that makes much more sense. 1 more question.

14 min later 5031005 Anonymous
>>5030991 Try converting tan to sin/cos.

19 min later 5031015 Anonymous
>>5030991 Ugh, I don't remember how to do this one without L'Hospital's rule or just plugging in numbers really close to 0 honestly. The sin/cos thing was my first instinct but I don't think that helps.

20 min later 5031018 Anonymous
>>5030991 I'm pretty sure (tan x/x) approaches 1, giving you 5/2. It may not be an identity but I used to do it

21 min later 5031020 BurnedHard_Reimann
>>5030991 Perhaps L'Hopital's rule may be of use. If \displaystyle \lim\limits_{x\to a}^{} \frac{\displaysyle f(x)}{\displaysyle g(x)} is indeterminant like 0/0, \infty / \infty etc, then \displaystyle \lim\limits_{x\to a}^{} \frac{\displaysyle f(x)}{\displaysyle g(x)}=\displaystyle \lim\limits_{x\to a}^{}\frac{\displaystyle f'(x)}{\displaystyle g'(x)} .

22 min later 5031023 Anonymous
>>5031018 If you use L'Hospital's Rule you get 5*sec^2(5x) / 2 sec(0) = 1/cos(0) = 1 5*1 /2 = 5/2

23 min later 5031024 BurnedHard_Reimann
>>5031020 mothafuckin' typos \displaystyle \lim\limits_{x\to a}^{}\frac{\displaystyle f(x)}{\displaystyle g(x)}=\displaystyle \lim\limits_{x\to a}^{}\frac{\displaystyle f'(x)}{\displaystyle g'(x)}

24 min later 5031026 Anonymous
>>5030991 >>5031018 >>5031020 >>5031023 it's 0. and i thought you guys knew math.

25 min later 5031032 Anonymous
>>5031026 oh wait, it's me who doesn't. it's 5/2

27 min later 5031036 Anonymous
OP here for the last time. I promise. If anyone can answer this, I'd really appreciate it. My teacher couldn't explain the concept b/c she doesn't speak English very well. Usually when you have a limit problem, you plug in what you approach to find the limit (once and if you can), as in the problem on the right. But, in the problem on the left, if you plug in pi you get 1/0, therefore the limit is 1/0. I understand conceptually why the limit on the left is infinity by looking at the graph and plugging in and infinitely small number, but the method is different from usual limits. Why is this?

27 min later 5031041 Anonymous
>>5031026 OP here, the answer is definitely 5/2 I have the answers

28 min later 5031043 Anonymous
>>5031036 ffs i forgot the picture again

29 min later 5031045 Anonymous (Untitled.png 1533x714 30kB)
>>5031043 omg

30 min later 5031049 Anonymous
>>5031036 you can only plug in the value you approach for continuous functions, or functions continuous on their domain.

42 min later 5031083 Anonymous
>>5031049 how do you know what is/isn't at first sight?

52 min later 5031101 Anonymous
>>5031083 Most often, division by 0.

55 min later 5031104 Anonymous
>>5031101 thanks, that's cleared up alot

59 min later 5031109 Anonymous
>>5031083 polynomials are continuous for all real numbers, you can always plug in a to find the lim of a polynomial as x approaches a. trig, inverse trig, root, exponential, logarithmic, and rational functions are continuous for every number in their domain. 1/sin(x) is a rational function but pi is not in it's domain, so you can't just plug in pi. In short, for any continuous function, if you can plug in the number x approaches and the answer is defined, that is the limit. If it is not, do some algebraic expansions/simplifications until f(a) is defined.

4 hours later 5031459 Anonymous
>>5030991 $\latex \lim_{x\rightarrow 0}\frac{\tan(5x)}{2x} =\lim_{x\rightarrow 0}\frac{5x+O((5x)^3)}{2x} =\lim_{x\rightarrow 0}\frac{5}{2}+\frac{O(125x^2)}{2} =\frac{5}{2} $

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