Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

L'Hopital's Rule Question?

Apply L'Hopital's Rule to evaluate the limit. It may be necessary to apply it more than once.

lim tan(3x)

x-> (pi/2) ---------

tan(5x)

1 Answer

Relevance
  • Hemant
    Lv 7
    1 decade ago
    Favorite Answer

    L = lim ( x → π/2 ) [ ( tan 3x ) / ( tan 5x ) ] ........... (1)

    .....................................................................................................................................................

    First, we note the following values :

    sin 3π/2 = -1, ... cos 3π/2 = 0 .......... tan 3π/2 = -∞

    sin 5π/2 = 1, ..... cos 5π/2 = 0 .......... tan 5π/2 = ∞

    ...........................................................................................

    Writing tan as sin/cos, and rearranging, the req'd limit is,

    L = lim ( x → π/2 ) [ ( sin 3x · cos 5x ) / ( sin 5x · cos 3x ) ]

    .. = { lim ( x → π/2 ) ( sin 3x / sin 5x ) } • { lim ( x → 0 ) ( cos 5x / cos 3x ) }

    .. = ( -1 / 1 ) • lim ( x → π/2 ) [ ( -5 sin 5x ) / ( -3 sin 3x ) ] ... by L' Hopital

    .. = ( -1 ) • [ -5(1) / ( -3 (-1) ) ]

    .. = ( -1 ) • ( - 5/3 )

    .. = ( 5/3 ) ...................................... Ans.

    ...........................................................................................

    Happy To Help !

    ...........................................................................................

Still have questions? Get your answers by asking now.