Callisto
  • Callisto
Find the limit:\[\lim_{x \rightarrow \infty} (\frac{x+2}{x-1})^x\]
Mathematics
katieb
  • katieb
See more answers at brainly.com
katieb
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Callisto
  • Callisto
Let \(y= (\frac{x+2}{x-1})^x\) \[ln y = x\ln(\frac{x+2}{x-1})\]Can't use L'Hopital's Rule now..Hmm..
AccessDenied
  • AccessDenied
I was thinking something like this: \( \displaystyle \frac{x + 2}{x - 1} = \frac{x - 1 + 3}{x - 1} = \frac{x - 1}{x - 1} + \frac{3}{x - 1} = 1 + \frac{3}{x-1}\) The end expression looks a lot more like the definition of e in terms of limits (lim as x appraoches infinity of (1 + 1/x)^x
zepdrix
  • zepdrix
You CAN apply L'Hop Rule Callisto, we just need to apply a tiny trick from the spot you got to. You want to write the x as the reciprocal of the reciprocal of x.\[\huge x\ln\left(\frac{x+2}{x-1}\right)=\frac{\ln\left(\frac{x+2}{x-1}\right)}{\frac{1}{x}}\] And you'll notice that from here, we get an indeterminate form, 0/0 which allows us to apply L'Hop.

Looking for something else?

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

More answers

Callisto
  • Callisto
@zepdrix My bad, I should have revised more... Thanks a ton!!! @AccessDenied I haven't thought of this way... :'( Perhaps I should give a try!!
zepdrix
  • zepdrix
Definition of e? :) Hmm that's clever, yah I like that.
AccessDenied
  • AccessDenied
Yea, I vaguely recall seeing it done somewhere, it seemed like it would be pretty useful here. :P
zepdrix
  • zepdrix
I remember it comes up if you need to use the Limit Definition of a Derivative to show the derivative of ln x. That one is kinda fun :3

Looking for something else?

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