Question

Question #62514

Answer 1

-4

Explanation:

best post only 1 question at a time

i can do the first one here

`lim_(x->1) ( x^2-1)/(sqrt(3x+1) - sqrt(5x-1))`

multiplying the denom by its conjugate is often a good idea when there are radicals present as is case here

`= lim_(x->1) ( x^2-1)/(sqrt(3x+1) - sqrt(5x-1)) * (sqrt(3x+1) + sqrt(5x-1))/(sqrt(3x+1) + sqrt(5x-1))`

`= lim_(x->1) ( x^2-1)/((3x+1) - (5x-1)) * (sqrt(3x+1) + sqrt(5x-1))`

`= lim_(x->1) ( x^2-1)/(-2x +2) * (sqrt(3x+1) + sqrt(5x-1))`

that right end of this limit is continuous through x = 1 so we can lift it out of the limit and apply its actual value

`= (sqrt(3x+1) + sqrt(5x-1)) lim_(x->1) ( (x-1)(x+1))/(-2(x-1))`

as we are still only looking at the limit, we can cancel the (x-1)'s

`= - 1/2 (sqrt 4 + sqrt 4) lim_(x->1) x+1`

and `lim_(x->1) x+1 = 2`

`\implies - 1/2 (4) 2 = -4`