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Math Help - L'Hopital Rule

  1. #1
    TWN
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    L'Hopital Rule

    1.) What is the limit of ( e^x - (1+x) ) / x^n as x approaches 0 from the right?

    Using the L'Hopital rule, I'm a little confused with regard to the denominator.

    2.) What is the limit of (squareroot(25-x^2)) / (x^2 - 1) when x approaches 5 from the left?

    Thank you in advance!
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    Re: L'Hopital Rule

    Hi!
    Fir the first one you need to apple L'Hospital recursively until you rid the 'x' term in the denominator

    (ex - (1+x)) / xn
    so differentiate to get
    (ex - 1 ) / n * xn-1
    Repeat
    (ex) / n(n-1) * xn-2

    So now you know that if you keep differentiating, the numerator remains unaffected. However the denominator will reduce to something like
    (n)(n-1)(n-2)......1 (= say N)

    Now put the limiting value of x=0
    you get
    1/ N


    Regarding the second one, i think the left hand limit will come out to be 0. Since the domain of the function is (-inf,5] , the LHL exists as x->5-

    Hope it helps !
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    Re: L'Hopital Rule

    Quote Originally Posted by TWN View Post
    1.) What is the limit of ( e^x - (1+x) ) / x^n as x approaches 0 from the right?

    Using the L'Hopital rule, I'm a little confused with regard to the denominator.

    2.) What is the limit of (squareroot(25-x^2)) / (x^2 - 1) when x approaches 5 from the left?

    Thank you in advance!
    Surely you know that the derivative of x^n is nx^(n-1)...
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    Re: L'Hopital Rule

    Quote Originally Posted by pratique21 View Post
    Hi!
    Fir the first one you need to apple L'Hospital recursively until you rid the 'x' term in the denominator

    (ex - (1+x)) / xn
    so differentiate to get
    (ex - 1 ) / n * xn-1
    Repeat
    (ex) / n(n-1) * xn-2

    So now you know that if you keep differentiating, the numerator remains unaffected. However the denominator will reduce to something like
    (n)(n-1)(n-2)......1 (= say N)

    Now put the limiting value of x=0
    you get
    1/ N


    Regarding the second one, i think the left hand limit will come out to be 0. Since the domain of the function is (-inf,5] , the LHL exists as x->5-

    Hope it helps !
    Isn't n(n-1)(n-2)...(3)(2)(1) = n! ?
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    Re: L'Hopital Rule

    Yeah. 1/n!
    Missed that. Thanks !
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    Re: L'Hopital Rule

    Quote Originally Posted by Prove It View Post
    Isn't n(n-1)(n-2)...(3)(2)(1) = n! ?
    Just btw, if we expand ex and then see, we have
    1+x+x2/2! + x3/3! + ...... xn/n! + ..... to inf

    so our expression becomes

    (1+x+x2/2! + x3/3! + ...... xn/n! + ..... to inf - (1+x) )/ xn
    = (x2/2! + x3/3! + ...... xn/n! + ..... to inf ) / xn


    now for terms beyond xn we need not worry as the higher powers will go to 0 as x-> 0
    however if we are to find the limit by dividing the rest of the terms :-
    (x2/2! + x3/3! + ...... xn/n!) / xn
    for x-> 0
    a number of undefined terms will be encountered, apart from 1/n! which seems to be coming if we can somehow deal with the lower powers !

    Kindly clarify this !
    thanks.
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