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Mathematics 12 Online
OpenStudy (anonymous):

limit of x is 1, 1/x-1/x-1 find the limit? do i start off by gettin a cd for 1/x-1?

OpenStudy (anonymous):

the answer is -1, how did the book get that answer?

OpenStudy (barrycarter):

Well, 1/x goes to 1 as x goes to 1, so 1/(x-1) is the only term that matters. It goes to infinity, so the entire limit goes to negative infinity.

OpenStudy (barrycarter):

Are you sure you typed it right? It's (1/x) - (1/(x-1)) ?

OpenStudy (barrycarter):

Or did you really mean 1/x - 1/x - 1 which simplifies to the constant -1 ?

OpenStudy (campbell_st):

rewrite it \[\lim_{x \rightarrow \1} ((x -1) - x)/x(x-1)\]

OpenStudy (anonymous):

\[\lim_{x \rightarrow 1}1/x-1/x-1\]

OpenStudy (barrycarter):

By order of operations, 1/x - 1/x - 1 is just -1.

OpenStudy (campbell_st):

oops... forget that treat it as

OpenStudy (campbell_st):

\[\lim_{x \rightarrow 1} 1/x - \lim_{x \rightarrow 1} 1/(x -1)\] limit of the 1st part is 1....

OpenStudy (campbell_st):

but 2nd part is undefinded

OpenStudy (anonymous):

righ, i believe that was the whole reason my professor gave us this problem, bcuz its 0/0 if u just plug in x, but i still dont get how the -1 comes out as an answer, like i need to be able to show on paper how i got -1

OpenStudy (anonymous):

its not undefined its indeterminent

OpenStudy (campbell_st):

well if it is 1/x - 1/x that is zero... and the constant term is -1

OpenStudy (anonymous):

Is that second1/x-1 term in the denominator too? or is it another term???

OpenStudy (precal):

I wonder if the student has been taught what to do with 0/0 yet?

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