Limit arctan(x) /x x->0?

Well you have not learned l'Hôpital's rule but how about learning it now. I have given you a link in the source which will tell you all about it.

We have zero over zero so find this limit with l'Hôpital's rule:

L = lim(x → 0) (tanˉ¹x / x)

L = lim(x → 0) [1 / (x² + 1)]

L = 1

There are at least two methods.

Method 1:

Let y = arctan(x). Then as x->0, y->0; also, x = tan(y). Therefore,

Limit x->0 arctan(x)/x

= Limit y->0 y/tan(y)

= Limit y->0 (y/sin(y))(sin(y)/tan(y))

= Limit y->0 (y/sin(y))(cos(y))

= Limit y->0 (cos(y)) / (sin(y)/y)

= [Limit y->0 cos(y)] / [Limit y->0 (sin(y)/y)]

= cos(0) / 1

= 1

Method 2: Since this limit has the indeterminate form 0/0, we can use L'Hopital's rule (differentiate numerator and denominator):

Limit x->0 arctan(x)/x

= Limit x->0 (1/(1 + x^2)) / 1

= (1/(1 + 0^2)) / 1

= 1

Lord bless you today!

Use L'Hopital's Rule.

The derivative of the numerator is: 1 / (1 + x^2)

If x = 0, that becomes 1 / (1 + 0^2) = 1 / (1 + 0) = 1/1 = 1

The derivative of the denominator is 1.

So the answer is: 1/1 = 1

You HAVE learned to differentiate.

By definition,

f'(a) = lim_(x -> a) (f(x) - f(a))/(x - a)

In this case, a = 0, f(x) = arctan(x), f'(x) = 1/(1+x^2); f'(0) = 1

limit = 1. The point of the problem is to recognize that the limit is a derivative.