Calculus

[Guest]

let F(x) = 5+5x^2 and g(x) = sqrt(x+4). find the limit

lim g(f(x))
x --> 3

[nathi123]

[tex]f(x)=5+5x^{2};g(x)=\sqrt{x+4}\Rightarrow\lim_{x \to 3}g(f(x))=\lim_{x \to 3}\sqrt{5+5x^{2}+4}=\sqrt{5+9.5+4}=\sqrt{54}=3\sqrt{6}[/tex].

[Guest]

All polynomials are continuous for all numbers and the square root function is continuous for all positive numbers. Additionally the composition of continuous numbers is continuous.

Now, do you know what "continuous" means?

A function, f(x), is "continuous" at x= a if and only if
i) f(a) exists.
ii) [tex]\lim_{x\to a} f(x)[/tex] exists.
iii) [tex]\lim_{x\to a} f(x)= f(a)[/tex].

That's why nathi could find the limit simply by evaluating the function.

Personally, I wouldn't have explicitly calculated g(f(x)).

I would have observed that, since [tex]f(x)= 5+ 5x^2[/tex], [tex]f(3)= 5+ 5(3^2)=5+ 45= 50[/tex] and, since [tex]g(x)= \sqrt{x+ 4}[/tex], [tex]g(f(3))= g(50)= \sqrt{50+ 4}= \sqrt{54}= \sqrt{9(6)}= 3\sqrt{6}[/tex].