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Mathematics 15 Online
OpenStudy (anonymous):

lim x→−5 5 − |x|/5 + x Find the limit of x Thank you for your help!

OpenStudy (anonymous):

(if it exists)

OpenStudy (anonymous):

Approach x = -5 from the left and from the right. If you are approaching the same value, the limit does exist. It might help to plot some points to see what's going on. The idea is that the limit can exist, even if the expression isn't solvable at that precise point

OpenStudy (anonymous):

Yes.. I do get that part, but I dont have a graph or anything?? Im not sure how I should approach it

OpenStudy (anonymous):

I keep getting 0 ..

OpenStudy (anonymous):

I haven't worked it, but the limit could be 0. Are you thinking that's wrong?

OpenStudy (anonymous):

Is the original problem a fraction where 5-|x| is in the numerator and (5+x) is the denominator?

OpenStudy (anonymous):

yes!

OpenStudy (anonymous):

I am plugging in the -5.. are you getting a zero as well?

OpenStudy (anonymous):

you can't just plug in -5, because the denominator would be zero and the function would be undefined

OpenStudy (anonymous):

oh..

OpenStudy (anonymous):

hmm..

OpenStudy (anonymous):

I GOT IT! it's 1

OpenStudy (anonymous):

the negative absolute value of x... is -x so.. haha yay

OpenStudy (anonymous):

If you sub in values for x like -10, -9, etc you always get f(x) = 1, right? And if you sub in values for x like 0, -1, -2, -3 etc you still get f(x) = 1 but at x = -5, it's undefined. However, the limit as you approach that point is 1 because the f(x) values on both sides, no matter how close you want to go, are still = 1

OpenStudy (anonymous):

good work... didn't see you already had it :)

OpenStudy (anonymous):

:) thanks!!!!

OpenStudy (anonymous):

thank you too!~

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