I am having some trouble with this particular type of question which asks,

Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval.

$$f(x) = x + \sqrt{x-4} \;\;\ , \;[\,4,\,\infty)$$

My attempt at the problem:

We know that a function is continuous at a point if

$$ \lim_{x\to a} f(x) = f(a)$$

and it is continuous on an interval if it is continous in every number in the interval. From the example problem in the textbook it seems that the first step is always to show the following:

$$\lim _{x\to a} f(x) = \lim_{x\to a} x + \sqrt{x - 4}$$ $$=\lim_{x\to a} x\,\,+ \lim_{x\to a}\sqrt{x-4}$$ $$=\lim_{x\to a}x \,\,+ \sqrt{\lim_{x\to a}x-4} $$ $$ = a + \sqrt{a-4}$$ $$= f(a)$$

What i understand of this is that we are essentially "creating" a function f(a) which we can easily use to show that certain limits of f(x) are equal to a corresponding value f(a) and thus prove continuity at that point.

Next i looked at the interval and tried to work out if f(x) is continuous at x=4, because our interval is from positive four to infinity so i reason if we can show that the limit as x approaches 4 from the right is equal to the function value at that point then we will know the function is continuous over that interval.

$$\lim_{x\to4^+} f(x) = 4 = f(4)$$

I'm not entirely sure of the last step - If i prove the function is right continuous at x=4 am i proving that every number in the interval is continuous because f(a) is a function and so as long as there are no restrictions we will get function values for every x that prove the function is continuous? What if the function had a denominator, how would i handle that situation? Any explanations for this type of problem would be greatly appreciated.

Thanks in advance


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