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How do you find the limit of #sintheta/(theta+tantheta)# as #theta->0#?

Answer 1

Paisley Johnson

#= 1/2#

#lim_(theta to 0) sintheta/(theta+tantheta)#

...is #0/0# indeterminate, so L'Hôpital's Rule appies...

# = lim_(theta to 0) (cos theta)/(1+sec^2 theta)#

# = lim_(theta to 0) (cos theta)/(1+1/(cos^2 theta))#

...all of which functions are continuous at the limit

# = lim_(theta to 0) (1)/(1+1/(1)) = 1/2#

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Answer 2

Elijah Abram

To find the limit of sin(theta)/(theta + tan(theta)) as theta approaches 0, we can use L'Hôpital's Rule. Taking the derivative of both the numerator and denominator with respect to theta, we get:

lim(theta->0) [cos(theta)/(1 + sec^2(theta))]

Substituting theta = 0 into the expression, we have:

lim(theta->0) [cos(0)/(1 + sec^2(0))]

Simplifying further:

lim(theta->0) [1/(1 + 1)]

Finally, evaluating the limit:

lim(theta->0) [1/2] = 1/2

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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 #sintheta/(theta+tantheta)# as #theta-&gt;0#?

How do you find the limit of #sintheta/(theta+tantheta)# as #theta->0#?