Calculus 1 explanation problem?

Question

I am supposed to explain what the equation Lim x->2 f(x) =5 means. What I think it means is that on a graph of y=5 ( a horizontal line) there is a limit at 2. Can someone confirm that or fix my logic in this? Thanks!

Answer 1

That takes me back. Can't remember. Good luck, you'll figure it out.

Answer 2

Can you explain to us what you just said? How can the horizontal line y=5 have a limit of 2? What does that have to do with f(x)?

Lim x->2 f(x) =5 means that as the values of x gets very close to 2, f(x) gets very close to 5.

Answer 3

Lim x->2 f(x) = 5; It means that as x approaches to the value "2" , the function f(x)approaches to a value of "5".

I think ur logic gets stuck if line, y =5, cuts or touches the function f(x) more than once.

Answer 4

Because the the function is continuous and constant,
the lim x-> a is 5 for all values of a including x - >2

Answer 5

In words this reads "As x-->2 , f(x)-->5."
Now y = 5 is a horizontal line that passes thro` (0,5)
So what we are now saying is that as x--> 2, f(x) approaches the line y = 5.
Line y = 5 is a horizontal asymptote.
f(x) may approach y = 5 from above or below the line.

Answer 6

it means that if you look at the graph y = f(x), when x gets very close to 2, y gets very close to 5. when x=2, y may not be equal to 5 or it may even be undefined. but when x is near 2, y is near 5.

Answer 7

f(a million) = 3 - ?(7) = approximately .3542 f(2) = approximately 20.sixty 8 for this reason the intermediate value theorem won't be able to be used. IVT demands a non-end equation, and it is. even regardless of the undeniable fact that a non-end equation could be drawn easily, and if we drew a line from .3542 to 20.sixty 8, it won't bypass in the time of the x axis. for this reason ivt gets you nowhere