How do you find the limit of #tan^-1(1/x)# as #x->0^+#?

Answer 1

Samantha Adkison

#lim_(x->0^+) arctan(1/x) = pi/2#

#lim_(x->0^+) arctan(1/x)#

substitute: #y= 1/x#, with #lim_(x->0^+) y(x) = +oo#

So:

#lim_(x->0^+) arctan(1/x) = lim_(y->+oo) arctan(y) = pi/2#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Emily Ammerman

To find the limit of tan^-1(1/x) as x approaches 0 from the positive side, we can use the properties of the inverse tangent function. As x approaches 0 from the positive side, 1/x approaches positive infinity. The inverse tangent function approaches π/2 as its argument approaches positive infinity. Therefore, the limit of tan^-1(1/x) as x approaches 0 from the positive side is π/2.

By signing up, you agree to our Terms of Service and Privacy Policy

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^-1(1/x)# as #x-&gt;0^+#?

How do you find the limit of #tan^-1(1/x)# as #x->0^+#?