Evaluate the limit of the function sin(3x)/tan(5x), if x goes to 0?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If we'll replace x by 0, we'll get the indetermination "0/0" type. Therefore, we can use L'Hospital's rule, to determine the limit of the quotient of derivatives.

lim f(x)/g(x) = lim f'(x)/g'(x)

lim sin(3x)/tan(5x) = lim [sin(3x)]'/[tan(5x)]'

We'll apply chain rule to differentiate with respect to x:

lim [sin(3x)]'/[tan(5x)]' = lim 3*cos (3x)/5/[cos (5x)]^2

We'll replace again x by 0 and we'll get:

lim 3*cos (3x)/5/[cos (5x)]^2 = 3*cos (0)/5/[cos (0)]^2

Since cos 0 = 1, we'll have:

lim 3*cos (3x)/5/[cos (5x)]^2 = 3/5

The requested limit of the function sin(3x)/tan(5x), when x approaches to 0, is: lim sin(3x)/tan(5x) = 3/5.

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