Question #62514

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Answer 1 · 2016-07-01T09:20:52

-4

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 $lim_(x->1) x+1 = 2$

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