anonymous
  • anonymous
lim x approaches 1 x-1/x^2+6x-7
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
is that x-1 on top of the rest? if so, try to use () to enclose the top and bottom parts...just makes it easier to read
amistre64
  • amistre64
(x-1)/(x^2+6x-7) as x->1 we of course get an indeterminate number 0/0
amistre64
  • amistre64
we should factor the bottom and see if that will elimiate some "problem" areas

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
on top
amistre64
  • amistre64
(x-1) --------- we see whats going on now (x+7)(x-1)
anonymous
  • anonymous
Exactly got the same thing by factoring
amistre64
  • amistre64
at x=1 there is a "hole" in the graph:
1 Attachment
amistre64
  • amistre64
so the limit actually "approaches" 1/8 if we cancel terms and apply the limit
anonymous
  • anonymous
Ah ok I follow.
amistre64
  • amistre64
the graphs for 1/(x+7) and the one you wrote are the same except for one point....x=1 has a "hole" in yours, that isnt there in this one....
anonymous
  • anonymous
Yeah I see that .

Looking for something else?

Not the answer you are looking for? Search for more explanations.