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How do you find the limit of #tan^2x/x# as #x->0#?

Answer 1

Charles Anctil

# lim_(x rarr 0)tan^2x/x = 0 #

# lim_(x rarr 0)tan^2x/x = lim_(x rarr 0)(tanxtanx)/x #

# :. lim_(x rarr 0)tan^2x/x = lim_(x rarr 0)sinx/cosx * sinx/cosx *1/x #

# :. lim_(x rarr 0)tan^2x/x = lim_(x rarr 0)sinx/x * 1/cos^2x * sinx #

# :. lim_(x rarr 0)tan^2x/x = lim_(x rarr 0)sinx/x * lim_(x rarr 0)1/cos^2x * lim_(x rarr 0)sinx #

# :. lim_(x rarr 0)tan^2x/x = lim_(x rarr 0)sinx/x * (1/(lim_(x rarr 0)cos^2x))^2 * lim_(x rarr 0)sinx #

# lim_(x rarr 0)sinx/x =0 # is standard calculus limit

# :. lim_(x rarr 0)tan^2x/x = 0 * 1 ^2* 0 = 0 #

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Answer 2

Aurora Alvarado

To find the limit of tan^2x/x as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of both the numerator and denominator, we get 2tan(x)*sec^2(x)/1. Evaluating this expression as x approaches 0, we find that the limit is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit of #tan^2x/x# as #x-&gt;0#?

How do you find the limit of #tan^2x/x# as #x->0#?