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Lemme write out the equation
and you wanna know if the function is continuous ?
hint : x^2-a^2 = (x-a)(x+a)
"In exercises 57-60, find the constant a, or the constants a and b, such that the function is continuous on the entire real line."
ok, hint : x^2-a^2 = (x-a)(x+a) helps ?
Sure but I don't know what to do after
you need to find the limit of g(x) when x->a
so see if the limit of (x^2-a^2)/(x-a) as x->a is equal to g(a) = 8
in order to find the limit of (x^2-a^2)/(x-a) as x->a use my hint.
[continuous means lim g(x) as x->a = g(a) ]
@neonpolarbears are you there ?
sorry i'm trying to work out the problem but it's just not making sense
have you calculated the limit of (x^2-a^2)/(x-a) as x->a ?
what did you get ?
does that get factored and cancelled?
yes x-a will cancel
(x^2-a^2)/(x-a) = (x-a)(x+a)/(x-a) = x+a what is the limit of x+a as x->a ?
so now we have lim g(x) as x->a = 2a and we need (for continuity) that this limit will be equal to g(a) which is 8. so what is a then for g(x) to be continuous ?
i'm so lost..
i have no idea how you got x to be approaching a
ok, continuous function at a point x=a satisfies the condition: lim g(x) as x->a = g(a) understand so far ?
this is the definition of continuous
tbh, constantly rewriting "the lim of so and so" just confuses me
is there a way to solve it without formally writing everything?
now lim g(x) as x->a = lim (x^2-a^2)/(x-a) as x->a = and as you said = 2a
this is what is mean when i write lim of something as x goes to something
i can understand it, just all the letters get overwhelming :/
im sorry for that :( i will write it
idk what you mean by 2a=g(a)=8
we found that limg(x) = 2a x->a right ? so now accroding to the definition for g(x) to be continuous limg(x) = g(a) x->a so we need 2a = g(a) but g(a) = 8 ( it is given in the question) so 2a = 8
then just solve for a?
sorry that took a long time - this is frustrating
but 2a=8 is not the important part as you know I hope you understand the procedure
if you have another question and you want to do it together ill be happy to help
(and im sorry for my english)