Limit without l'hopitals rule?

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Using the DEFINITION OF DIFFERENTIATION:

Let f(x) = arctan(x).

lim_(h -->0)) f(h)/h = lim_(h --> 0) (f(h) - f(0))/h = f'(0)

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

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L'Hospital's rule came long after this historically & in your textbook.
It is 100% true that I did not use it.

However, another way is this:
Let y = arctan(x), so x = tan(y)

Then want lim _(y --> 0) y/tan(y)

tan(y)/y = (1/cos y)(sin(y)/y)

cos(y) --> 1 as y --> 0

sin(y)/y --> 1 as y --> 0. This has some sort of geometric proof at the beginning
of the study of differentiating trig functions.

Once again, limit = 1....Show more

Well you can simplify a bit. Let theta = arctan(x), and so

1/sqrt(1+x^2) = cos(theta)

or

x = sign(theta) sqrt(1/cos^2(theta) -1),

(sign(theta) = 1 if theta>0, and -1 if theta< 0) and so you can change the limit to

limit theta-> 0 |theta|/sqrt(1/cos^2(theta) -1) = limit theta-> 0 theta*cos(theta)/sin(theta)

This simplifies the problem to calculating

limit theta-> 0 theta/sin(theta),

and maybe you already know that, or know how to do it ......Show more

you ought to use the enlargement of cos x to sparkling up this subject. cos(x) = a million - x^2/2! + x^4/4! - x^6/6! + ... So, a million - cos(x) = x^2/2! - x^4/4! + x^6/6! + ... (a million-cos(x))/x^2 = a million/2! - x^2/4! + x^4/6! + ... restricting x to 0, we get a million/2 using fact the respond...Show more