calculus question..?

Question

how does..

[ln (e - 1 + 1)] / [(e-1) +1] = 1/e ..? what cancels out and how or why..?

also, how do you go from...
[1/t + 1] / [2(ln t + t)^1/2]

to... [1+t] / 2t (ln t + t)^1/2 ..?

Answer 1

Simplify.

-1 + 1 = 0 in both the numerator and denominator.

[ln (e - 1 + 1)] / [(e - 1) + 1] = ln(e) / e = 1/e
______________

[1/t + 1] / [2(ln t + t)^(1/2)]

Multiply numerator and denominator by t.

= [1 + t] / [2t(ln t + t)^(1/2)]

That's where you stopped but I would continue to simplify.

= [1 + t] / [t(ln t + t)^(2/2)]

= [1 + t] / [t(ln t + t)]
____________

Answer 2

e - 1 + 1 = e, so remembering to evaluate inside brackets first, the first expression reduces to
ln e / e
and ln e = 1 by definition, so we get 1/e.

For the second one, we are just multiplying top and bottom by t. (Note that t(1/t + 1) = t(1/t) + t = t/t + t = 1 + t.) The extra t on the bottom is just after the 2, if you missed it.