survey of clac

posted by .

g(x)={x+6, for x<-2,
-1/2x+1, for x> or equal to -2,

find the limit:
lim g(x)= lim g(x)=
x-->-2^- x-->-2^+

lim g(x)= lim g(x)=
x-->-2^+ x-->4^-

  • survey of clac -

    CLAC as the School Subject is not helpful.

    Sra

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. calc

    need to find: lim as x -> 0 of 4(e^2x - 1) / (e^x -1) Try splitting the limit for the numerator and denominator lim lim x->0 4(e^2x-1) (4)x->0 (e^2x-1) ______________ = ________________ lim lim x->0 e^X-1 x->0 e^x-1 …
  2. math

    i need some serious help with limits in pre-calc. here are a few questions that i really do not understand. 1. Evaluate: lim (3x^3-2x^2+5) x--> -1 2. Evaluate: lim [ln(4x+1) x-->2 3. Evaluate: lim[cos(pi x/3)] x-->2 4. Evaluate: …
  3. calc bc (condensed

    is the limit as x approaches 0 of sin3x over 3x equal to zero?
  4. Calc. Limits

    Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity?
  5. calc

    Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity?
  6. calculus

    f(x)= 3x+2, x<0 3, x=0 x^2+1, x>0 g(x)= -x+3, x<1 0, x=1 x^2, x>1 find a) lim x->0+ g(f(x)) b)lim x->0- g(f(x)) c) lim x->1+ f(g(x)) d)lim x->1- f(g(x)) I am trying to understand these. Help appreciated. Have …
  7. Calculus

    Find the indicated limits. If the limit does not exist, so state, or use the symbol + ∞ or - ∞. f(x) = { 2 - x if x ≤ 3 { -1 + 3x - x^2 if x > 3 a) lim 3+ f(x) x->3 b) lim 3- f(x) x->3 c) lim f(x) x->3 …
  8. Calculus

    Find the indicated limits. If the limit does not exist, so state, or use the symbol + ∞ or - ∞. f(x) = { 2 - x if x ≤ 3 { -1 + 3x - x^2 if x > 3 a) lim 3+ f(x) x->3 b) lim 3- f(x) x->3 c) lim f(x) x->3 …
  9. Calculus

    Find the following limits algebraically or explain why they don’t exist. lim x->0 sin5x/2x lim x->0 1-cosx/x lim x->7 |x-7|/x-7 lim x->7 (/x+2)-3/x-7 lim h->0 (2+h)^3-8/h lim t->0 1/t - 1/t^2+t
  10. Calculus

    Find the limit. lim 5-x/(x^2-25) x-->5 Here is the work I have so far: lim 5-x/(x^2-25) = lim 5-x/(x-5)(x+5) x-->5 x-->5 lim (1/x+5) = lim 1/10 x-->5 x-->5 I just wanted to double check with someone and see if the answer …

More Similar Questions