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# 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)

Relevance
• Hemant
Lv 7

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

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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 = ∞

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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.

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Happy To Help !

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