You may also approach a method of solving,l'Hospital's theorem, commonly used in case of `0/0 ` indetermination.

`lim_(y-gt4) (5-(y^2+9)^(1/2))/(y-4) = lim_(y-gt4) ((5-(y^2+9)^(1/2))')/((y-4)')`

`lim_(y-gt4) ((5-(y^2+9)^(1/2))')/((y-4)') = lim_(y-gt4) -((2y)/(2sqrt(y^2+9)))/1`

`lim_(y-gt4) ((5-(y^2+9)^(1/2))')/((y-4)') = lim_(y-gt4) -((y)/(sqrt(y^2+9)))`

Substituting 4 for y yields:

`lim_(y-gt4) -((y)/(sqrt(y^2+9))) = -((4)/(sqrt(4^2+9)))`

`lim_(y-gt4) -((y)/(sqrt(y^2+9))) = -4(sqrt(25))`

`lim_(y-gt4) -((y)/(sqrt(y^2+9))) = -4/5`

**Hence, evaluating the limit yields `lim_(y-gt4) (5-(y^2+9)^(1/2))/(y-4) = -4/5.` **

`lim_(y-gt4)(5-sqrt(y^2+9))/(y-4)`

Now if you try to substitute y =4, and try to calculate the limit straight away, you would get 0/0 which is not definite value.

So we will modify the expression to remove this from happenning.

we can multiply both numerator and denominator by `(5+sqrt(y^2+9))`

`lim_(y-gt4)((5-sqrt(y^2+9))*(5+sqrt(y^2+9)))/((y-4)*(5+sqrt(y^2+9)))`

Now the numerator can be simplified into,

`lim_(y-gt4)(25-(y^2+9))/((y-4)*(5+sqrt(y^2+9)))`

`lim_(y-gt4)(-(y^2-16))/((y-4)*(5+sqrt(y^2+9)))`

`lim_(y-gt4)(-(y-4)(y+4))/((y-4)*(5+sqrt(y^2+9)))`

`lim_(y-gt4)(-(y+4))/(5+sqrt(y^2+9))`

now we can apply y =4 and calculate the limit.

`lim_(y-gt4)(-(4+4))/(5+sqrt(4^2+9)) = (-8)/(5+sqrt(25)) = (-8)/(5+5) = (-8)/10 = (-4)/5`

Therefore,

`lim_(y-gt4)(5-sqrt(y^2+9))/(y-4) = (-4)/5`