Use the definition of limit and properties of limit to show that function is continuous at the given number
f(x)=x2+√7-x , a=4

Question

Use the definition of limit and properties of limit to show that function is continuous at the given number
f(x)=x2+√7-x           , a=4

anonymous · Answer

I'm not sure whether I've read the function in the right way, but I think you mean $f(x) =x^2 + \sqrt{7-x}$ and not $f(x) = 2x + \sqrt{7} -x$

If it's correct, you can see that $f(x)$ is composed of 2 terms having x as a variable. So you can view each of them as an independent function: $x^2$ and $\sqrt{7-x}$. This will help us to see how to apply the properties of the limits.

By definition, $f(x)$ will be continuous at $a=4$ if $\lim_{x \to a} f(x)$ exists and  it's equal to $f(a)$. On the left-hand side we're going to prove that in the faster way, by plugging 4 in $f(x)$ and on the left we're going to do the same but by using the limit properties. Let use 4 instead of a:

$$\large{\lim_{x \to 4} (x^2 + \sqrt{7-x}) = \lim_{x \to 4} (x^2 + \sqrt{7-x})}$$
$$\large{4^2 + \sqrt{7-4} = \lim_{x \to 4} (x^2) + \lim_{x \to 4}(\sqrt{7-x})\hspace{35pt} (1)}$$
$$\large{16 + \sqrt{3} = \left(\lim_{x \to 4} x\right)^2 + \sqrt{\lim_{x \to 4}(7-x)}\hspace{35pt} (2)}$$
$$\large{16 + \sqrt{3} = 16 + \sqrt{7-4}}\hspace{35pt} (3)$$
$$\large{16 + \sqrt{3} = 16 + \sqrt{3}}\hspace{35pt} (4)$$

$(1)$: Take the limit as if it were a sum of two functions.
$(2)$: The property for functions rise to some number: the first to the 2nd power and the second to the $\frac{1}{2}$ power.
$(3)$: Just arithmetic.
$(4)$: We're done.