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Math Help - L'Hopital's rule

  1. #1
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    L'Hopital's rule

    Hi,
    Why can't i use L'Hopital's rule in this limit:

    lim (2x+sin2x+1)/(2x+sin2x)(sinx+3)^2 as x->infinity

    Thanks...
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  2. #2
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    Quote Originally Posted by Tom92 View Post
    Hi,
    Why can't i use L'Hopital's rule in this limit:

    lim (2x+sin2x+1)/(2x+sin2x)(sinx+3)^2 as x->infinity

    Thanks...
    I assume this is

    \lim_{x \to \infty}\frac{2x + \sin{2x} + 1}{(2x + \sin{2x})(\sin{x} + 3)^2}.

    You can only use L'Hospital's Rule if you get \frac{0}{0} or \frac{\infty}{\infty} from direct substitution.

    But in this case, you'll get \frac{1}{0}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    I assume this is

    \lim_{x \to \infty}\frac{2x + \sin{2x} + 1}{(2x + \sin{2x})(\sin{x} + 3)^2}.

    You can only use L'Hospital's Rule if you get \frac{0}{0} or \frac{\infty}{\infty} from direct substitution.

    But in this case, you'll get \frac{1}{0}.
    Dear Prove It,

    I think you have made mistake in this case. Note that the limit goes to "infinity". Hence the denominator and numerator both tends to "infinity."
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  4. #4
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    Quote Originally Posted by Sudharaka View Post
    Dear Prove It,

    I think you have made mistake in this case. Note that the limit goes to "infinity". Hence the denominator and numerator both tends to "infinity."
    Oh, oops. Hahaha, I subsituted x = 0 instead.

    Then I don't see why you can't use L'Hospital's Rule in this case...
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    Quote Originally Posted by Prove It View Post
    Oh, oops. Hahaha, I subsituted x = 0 instead.

    Then I don't see why you can't use L'Hospital's Rule in this case...
    Dear Prove It,

    But notice that,

    \lim_{x\rightarrow{\infty}}\frac{2x+\sin{2x}+1}{(2  x+\sin{2x})(\sin{x}+3)^2}

    Dividing the numerator and denominator by "2x";

    \lim_{x\rightarrow{\infty}}\frac{1+\frac{\sin{2x}}  {2x}+\frac{1}{2x}}{(1+\frac{\sin{2x}}{2x})(sinx+3)  ^2}

    Therefore the limit does not exist.
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