- #1

singedang2

- 26

- 0

[tex]\lim_{x\rightarrow\infty}\frac{\sin|x|}{x}[/tex]

[tex]\lim_{x\rightarrow 0}\frac{|x|-|x-2|}{x-1}[/tex]

thanks!

[tex]\lim_{x\rightarrow 0}\frac{|x|-|x-2|}{x-1}[/tex]

thanks!

Last edited:

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- Thread starter singedang2
- Start date

- #1

singedang2

- 26

- 0

[tex]\lim_{x\rightarrow 0}\frac{|x|-|x-2|}{x-1}[/tex]

thanks!

Last edited:

- #2

Gib Z

Homework Helper

- 3,352

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-1/infinity=-0=0, anything inbetween will also equal zero, except zero it self, but as u can see, the limit as it approaches that is also zero, i know im not very clear..ok basically its zero..and the 2nd one, a little easier :D well basically people use limits to find out wat the value something wud approach at an asymtote or a discontinous point. however, in this case, there is no discontinuity at point zero, so you can just sub it straight in and you'll be fine. so its (0 - l0-2l)/(0-1)= (0- +2)/-1= 2

i know my layout is bad, i hope u get da gist of it..

- #3

HallsofIvy

Science Advisor

Homework Helper

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For the second, do the two onesided limits: For x close to 0 and x> 0, |x|= x and |x- 2|= 2- x ( x is close to 0 so x< 2, x- 2 is negative). For x close to 0 and x< 0, |x|= -x and |x-2|= 2- x (x is still less than 2).

(Gib Z is correct- since the second function is continuous at x= 0, you can just substitute x= 0 but it might be that you want to find the limit in order to show that the function is continuous.)

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