how can i solve this limit problem?

Question

lim x^2 + 9/x^2 - 1
x->4

Answer 1

In general when you are taking a limit of a function f(x) where f(x) is defined at a and is continuous around a then the limit is f(a)

That is

lim x -> a f(x) = f(a) ; if f(a) exists and is continuous around a

note that f(x) = x^2 + 9/x^2 -1 is continuous and defined for all x but x = 0

So lim x -> 4 f(x) = f(4) = 4^2 +9/4^2-1 = 16+9/16-1 = 15+9/16 = 231/16

Answer 2

Since all the expressions are defined for x=4 and the bottom isn't 0 when x=4, it is ok to just plug the value in.

(4^2 + 9)/(4^2 -1) = 25/15 = 5/3.

Answer 3

The limit of f = x² + 9/(x²-1) at x = 4 is, by definition

f(4+h) as h--> 0

so

(4 + h)² + 9/((4+h)² - 1)

expanding

16 + 8h + h² + 9/(16 + 8h +h² -1)

letting h go to 0

16 + 9/(16-1) = 16.6


Doug

Answer 4

In the problem given, just substitute x = 4 and evaluate it -- there is no division by zero or other problems. If some denominator went to zero, one would use l'Hopital's rule: differentiate with respect to x, and evaluate the derivative at x = 4.

Answer 5

Just plug in 4 for x and you get 16.6