anonymous
  • anonymous
find the limit of (1/t-1/t^2+t) as t approaches 0
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jdoe0001
  • jdoe0001
\(\bf lim_{t \rightarrow 0} \left( \cfrac{ \frac{1}{t-1} }{ t^2+t } \right) \quad ?\)
jdoe0001
  • jdoe0001
hmm, could be yes
anonymous
  • anonymous
my computer is being slow sorry and it's set up like the other guy had it

Looking for something else?

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

More answers

jdoe0001
  • jdoe0001
who is "da odher guy"?
anonymous
  • anonymous
|dw:1378408411700:dw|
jdoe0001
  • jdoe0001
well, just add them up, and you'get your limit :)
anonymous
  • anonymous
don't you have to do something so you can add them??
jdoe0001
  • jdoe0001
well, yes, get LCD for the denominators
jdoe0001
  • jdoe0001
so your denominators are \(\bf t \ and \ t^2+t\)
jdoe0001
  • jdoe0001
keep in mind that \(\bf t^2+t \implies t(t+1)\) taking common factor
jdoe0001
  • jdoe0001
so, what do you think would be the LCD for the factors of \(\bf t \quad, \quad t(t+1)\)
anonymous
  • anonymous
I don't quite understand your terminology (quad, bf)
jdoe0001
  • jdoe0001
hehe, ohh man, you're using Internet Explorer.... try the same page in firefox or chrome :|
jdoe0001
  • jdoe0001
or even opera
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
ok the common factor is t, correct?
jdoe0001
  • jdoe0001
well, yes, the common factor is "t" but what you want is the LCD, not the common factor so you want the "common dividend" really
jdoe0001
  • jdoe0001
usually depending on the expressions, if they don't factor it simple enough, you can always just multiply them to each other and divide from there
jdoe0001
  • jdoe0001
notice that in \(\bf t \quad, \quad t(t+1)\) t(t+1) contains "t" and t(t+1) is divided by itself too thus the LCD will be the t(t+1)
jdoe0001
  • jdoe0001
\(\bf lim_{t \rightarrow 0} \left( \cfrac{1}{t} - \cfrac{1}{t^2+t} \right)\\ \quad \\ \cfrac{1}{t} - \cfrac{1}{t^2+t} \implies \cfrac{\square? - \square?}{t(t+1)}\)
anonymous
  • anonymous
t+1, 1
anonymous
  • anonymous
the limit is 1??
anonymous
  • anonymous
i mean 0
jdoe0001
  • jdoe0001
hmmm, one sec
jdoe0001
  • jdoe0001
hmm, what did you get for the numerator?
anonymous
  • anonymous
t
jdoe0001
  • jdoe0001
ok then \(\bf lim_{t \rightarrow 0} \left( \cfrac{1}{t} - \cfrac{1}{t^2+t} \right)\\ \cfrac{1}{t} - \cfrac{1}{t^2+t} \implies \cfrac{(t+1)-1}{t(t+1)} \implies \cfrac{\cancel{t}}{\cancel{t}(t+1)}\)
jdoe0001
  • jdoe0001
what would that give you?
anonymous
  • anonymous
1
jdoe0001
  • jdoe0001
:)
anonymous
  • anonymous
another question... if the x you are approaching would make both the numerator and denominator 0 can you use direct substitution to make 0 over 0=1??
jdoe0001
  • jdoe0001
well, that's the eternal quandary hehe 0/0 is not 1, is just indeterminate
jdoe0001
  • jdoe0001
http://www.mathsisfun.com/numbers/dividing-by-zero.html
anonymous
  • anonymous
thanks!
jdoe0001
  • jdoe0001
yw

Looking for something else?

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