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Babynini
 one year ago
More limits :)
Babynini
 one year ago
More limits :)

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[ x = \begin{cases} x \ \text{ if } \ x \ge 0\\ x \ \text{ if } \ x < 0\\ \end{cases} \] \[ \color{red}{x} = \begin{cases} (\color{red}{x}) \ \text{ if } \ (\color{red}{x}) \ge 0\\ (\color{red}{x}) \ \text{ if } \ (\color{red}{x}) < 0\\ \end{cases} \] \[ \color{red}{2x+1} = \begin{cases} (\color{red}{2x+1}) \ \text{ if } \ (\color{red}{2x+1}) \ge 0\\ (\color{red}{2x+1}) \ \text{ if } \ (\color{red}{2x+1}) < 0\\ \end{cases} \] \[ 2x+1 = \begin{cases} 2x+1 \ \text{ if } \ x \ge \frac{1}{2}\\ 2x1 \ \text{ if } \ x < \frac{1}{2}\\ \end{cases} \] based on the last definition above, we can replace the `2x+1` with just `2x+1` because x is getting closer and closer to 0 (0 makes x >= 1/2 true)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2Similarly, \[ 2x1 = \begin{cases} 2x1 \ \text{ if } \ x \ge \frac{1}{2}\\ 2x+1 \ \text{ if } \ x < \frac{1}{2}\\ \end{cases} \] So `2x1` can be replaced with `2x+1` (x=0 makes x < 1/2 true)

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0How did you come up with the 1/2 's ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2solving 2x+1 >= 0, 2x+1 < 0, etc etc. The restrictions placed on each piece of the piecewise function

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0So now we replace the values in the original with the ones you've just come up with.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0So it's [(2x+1)(2+1)]/x

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2close, there's an x missing

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Right on the second 2, sorry xD

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0But that al simplifies to 4x/x?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2Sorry you should have this \[ \lim_{x\to 0} \frac{2x12x+1}{x} = \lim_{x\to 0} \frac{(2x+1)(2x+1)}{x} \]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you somehow mixed up 2x1 and 2x+1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2It should be 4. The graph confirms it https://www.desmos.com/calculator/4uqoaaryhi Definitely an odd and interesting graph

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0okay, I got it right now :) so the entire limit =  4

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2correct, as x gets closer and closer to 0 (from either side) the value of y gets closer and closer to 4

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Fabulous. Thank you so much. So at the end I would just rewrite the original limit and then put it all equals 4? correct?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2yeah after you simplify you'll get 4x/x = 4 then you just apply the limit \[\Large \lim_{x\to 0}(4)\] to get 4 itself
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