I have the following statement:
Prove if $\sqrt{log(x^2+7)}$ is continuous at $x=-4$
My development was:
Let $g(x)=\log(x)$ and $ f(x)=x^2+7$
I will prove that $\log(x^2+7)$ is continuous at $x=-4$.
$\log(x^2 +7)$ is continuous at $x = -4 \iff \lim_{x\to -4}g(f(x)) = g(f(-4))$
Since $f$ is a polynomial is continuous at $-4 \in \mathbb{R}$
Also, $g(x)$ is continuous in $\mathbb{R^+}$ hence is continuos at $f(-4) = 23 \in \mathbb{R^+}$
$(\star)$With these conditions, i can say that $\lim_{x\to -4}g(f(x)) = g(f(-4))$ is true, i.e, $(g\circ f)(x)$ is continuous at $x = -4$.
Let $h(x) = \sqrt{x}$ and $p(x)= g(f(x)) = \log(x^2 + 7)$
Since $h$ is continuous in $\mathbb{R^+_0}$ and $p(-4) = log(23) \in \mathbb{R^+_0},$ hence $h$ is continuous in $p(-4)$. Furthermore, using $(\star)$ i can affirm that $\lim_{x\to -4}h(p(x)) = h(p(x)) = \sqrt{log(23)}$, that is $h(g(f(x))) = \sqrt{\log(x^2+7)}$ is continuous at $x-4$
Is my development correct?
Thanks in advance for take your time for reading.
Your method is absolutely correct. Nicely done! I, however, like approaching this via the $\epsilon$ method. We note:
$$\lim_{x\to a}f(x)=f(a)\iff \lim_{\epsilon\to0}f(a\pm\epsilon)=f(a)$$
such that $\epsilon>0$
Here: $$\lim_{\epsilon\to0} f(-4\pm\epsilon)=\lim_{\epsilon\to0}\sqrt{\log((-4\pm\epsilon)^2+7)}=\lim_{\epsilon\to0}\sqrt{\log(23+\epsilon^2\mp8\epsilon)}=\sqrt{\log(23)}=f(-4)$$