1. Questions

calculus

Find the value of the limit.( use L'hospital rule)

lim (tanx/x)^(1/x^2)
x>0

  1. 👍
  2. 👎
  3. 👁
Yen

  1. In consideration of the power (1/x^2), we can evaluate using l'Hopital's rule:
    k = limit of the logarithm of the given expression

    The limit of the given expression will then be L = ek

    k
    =lim(x->0){(log(tan(x))-log(x))/x²}

    Since we end up in a fraction where both numerator and denominator evaluate to zero when x->0, we can use l'Hopital's rule.

    We differentiate the numerator to get:
    N1=sec²(x)/tan(x) - 1/x
    which evaluates to ∞-∞ as x->0
    Differentiate the denominator to get
    D1=2x which evaluates to zero as x->0

    Repeat the same exercise to get:
    N2=2sec²(x) + 1/x^sup2 - sec4(x)/tan²(x)
    As x->0, 2sec²(x) evaluates to 2
    while
    1/x^sup2 - sec4(x)/tan²(x) evaluates to ∞-∞.

    Express the latter expression as:
    1/x^sup2 - sec4(x)/tan²(x)
    = (x²sec²(x)-tan²(x)) / tan(x)x²
    and evaluate again by l'hopital's rule:
    n1=(2sec(x)^2-4x^2sec(x)^4)tan(x) - 2xsec(x)^4
    d1=2xtan(x)^2+2x^2sec(x)^2tan(x)
    Both evaluate to zero, so we apply l'Hopital's rule again until both evaluate to a finite value, in fact,
    n4=-32, and d4=24
    Therefore
    N2 = 2 - 32/24 = 2/3

    The denominator becomes 2 after differentiation:
    D2=2

    Thus the limit of the logarithm of the given expression is
    N2/D2=(2/3)/2 = 1/3

    The limit of the given expression can be obtained by taking the anti-logarithm of 1/3, namely
    e(1/3)

    Check my work.

    1. 👍
    2. 👎
    MathMate

  2. Hi, that's the correct answer but i couldn't follow with what you did. what is sup means? and what's n1 and N1. Could you please elaborate more in details? thanks alot.

    1. 👍
    2. 👎
    Yen

Respond to this Question

First Name

Your Response

Similar Questions

  1. calculus

    Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→0 (e^x − e^−x − 2x)/(x − sin(x))

  2. calculus

    Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→∞ {(5x − 4)/(5x + 3)}^5x + 1

  3. Calculus

    Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim (tan(7x))^x x~>0

  4. Calculus Limits

    Question: If lim(f(x)/x)=-5 as x approaches 0, then lim(x^2(f(-1/x^2))) as x approaches infinity is equal to (a) 5 (b) -5 (c) -infinity (d) 1/5 (e) none of these The answer key says (a) 5. So this is what I know: Since

  1. Calculus

    find the limit. use L'Hopital's Rule if necessary. lim (x^2+3x+2)/(x^2+1) x -> -1

  2. calculus

    1) find the indicated limit, if it exist? a) lim x->-2 (x^2 -9)/(x^2+x-2) b) lim x -> -∞ √(ax^2+bx+c)/dx + e, where a > 0, b,c,d, and e are constant.

  3. 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 is

  4. Calculus

    Find the limit lim as x approaches (pi/2) e^(tanx) I have the answer to be zero: t = tanx lim as t approaches negative infi e^t = 0 Why is tan (pi/2) approaching negative infinity is my question?

  1. Calculus

    Evaluate the limit using L'Hospital's rule if necessary. lim as x goes to +infinity x^(6/x)

  2. Calc 1

    Use l'Hospital's Rule to find the exact value of the limit. lim x→∞ (1+4/x)^x

  3. Calc 1

    Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→3 (x^2+3x−18)/(x−3)

  4. calculus 1

    Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. lim x → 0 (x 6x)/(6x − 1)

You can view more similar questions or ask a new question.