Simple calc limit problem help

  • Thread starter Thread starter Matus1976
  • Start date Start date
  • Tags Tags
    Limit
Matus1976
Messages
4
Reaction score
0
I'm in Calc I and have run into a probably simple problem I can use guidance on.

Evaluate the following Limit

lim x -> -1 [(2x-1)^2 - 9] / [x+1]

Its been a few years since pre-calc and I'm drawing a blank on how to proceed on this problem. Thanks for any help offered!
 
Physics news on Phys.org
Try to factor (2x-1)^2-9 and see if x+1 is a factor. I think it is.
 
That's what it was, thanks!

How about this one?? The book has the answer as -1

lim x -> b [(x-b)^50 - x + b] / [x - b]
 
Matus1976 said:
That's what it was, thanks!

How about this one?? The book has the answer as -1

lim x -> b [(x-b)^50 - x + b] / [x - b]

Write the numerator as (x-b)^50-(x-b). It is divisible by (x-b), right?
 
Dick said:
Write the numerator as (x-b)^50-(x-b). It is divisible by (x-b), right?

Yeah that did it, thanks again! I didn't think to factor the -1 out.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top