Calculus

posted by .

how do you find the limit at infinity of:

lim(x->infinity) (x+2)/sqrt(64 x^2+1)

Do you first change the square root on denominator to (64x^2+1)^-1/2 and then divide everything by zero.

Please help me with this, I'm confused.

  • Calculus -

    Divide by zero? That is not allowed here in Texas.


    expand the denominator by the binomial expansion...


    (64x^2+1)^.5 = (64x^2)^.5+ Cn,k(64x^2)^-.5+Cn,k(64x^2)^-1.5+....

    well, as x>inf, all the terms after the first go to zero,
    So the denominator becomes 8x

    Limx>inf (x+2)/(8x+ ...)= 1/8

    One other way. Multipy numerator and denominator by 1/x

    (1+2/x)/sqrt(64+1/x^2)

    now, as the limit as x>large then is
    1/sqrt64= 1/8

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus - ratio test

    infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] | = lim (n->infinity) | ((e^n)(e^1)(n!)) …
  2. calculus - ratio test

    Posted by COFFEE on Sunday, July 29, 2007 at 6:32pm. infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] …
  3. calculus - ratio test

    infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] | = lim (n->infinity) | ((e^n)(e^1)(n!)) …
  4. Algebra

    What is the answer? I came up with C. is this correct?
  5. Calc. Limits

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

    Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity?
  7. Calc Please Help

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

    Find the horizontal asymptote of f(x)=e^x - x lim x->infinity (e^x)-x= infinity when it's going towards infinity, shouldn't it equal to negative infinity, since 0-infinity = - infinity lim x-> -infinity (e^x)-x= infinity
  9. Check my CALCULUS work, please! :)

    Question 1. lim h->0(sqrt 49+h-7)/h = 14 1/14*** 0 7 -1/7 Question 2. lim x->infinity(12+x-3x^2)/(x^2-4)= -3*** -2 0 2 3 Question 3. lim x->infinity (5x^3+x^7)/(e^x)= infinity*** 0 -1 3 Question 4. Given that: x 6.8 6.9 6.99 …
  10. Math- Calculus

    I need help with these problems, I cannot find a similar example to help me in the book: 1. Find lim x->infinity (e^(-2x) + sin x). 2. Find the derivative of sqrt(9-x) using the limit process. 3. Find lim x-> -infinity (x + sqrt(x^2 …

More Similar Questions