Question

lim as x=>3- ((sqrtx^2+16)-5)/(x^2-9)

Answer 1

rationalise the numerator by multiplying both numerator and denominator by sqrt(x^2+16) +5 you get
( x^2 +16 - 25)/((x^2-9)(sqrt(x^2+16)+5))
hence you get x^2-9 in both numerator and denominator
so the limit simplifies to 1/(sqrt(x^2+16)+5)
now x=3
so the limit is 1/10.

Answer 2

sorry, the x^2-9 is also in a sqrt

Answer 3

ok could you please simplify the question what abt the -5 and the +16?

Answer 4

ok it's \[(\sqrt{x^2+16}-5)/\sqrt{x^2-9}\]

Answer 5

Then the limit is just 0