Question

Plz with steps .. it's very hard can anyone plz

Answer 1

Explanation:

`\displaystyle\ \lim_(n \to a) (√(2x)-√(3x-a) )/(√(x)-√(a)) =(0)/(0) \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=√(2x)-√(3x-a) \qquad f'(x)=(2)/(2*√(2x)) -(3)/(2*√(3x-a)) \\\\g(x)=√(x) -√(a) \qquad g'(x)=(1)/(2√(x)) \\\\\\\displaystyle\ \lim_(n \to a) (√(2x)-√(3x-a) )/(√(x)-√(a)) =\lim_(n \to a) ((2)/(2*√(2x)) -(3)/(2*√(3x-a)) )/((1)/(2√(x)) )\\\\`

`\displaystyle \lim_(n \to a) (2√(x) )/(√(2x)) -(3*√(x) )/(√(3x-a)) =\lim_(n \to a) (2 )/(√(2)) -(3*√(x) )/(√(3x-a))\\\\\\=(2)/(√(2)) -(3*√(a) )/(√(2a))\\\\\\=(2)/(√(2)) -(3)/(√(2))\\\\\\=-\ (1)/(√(2))\\\\`