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

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

This question aims to explain the concepts of continuity in functions, the difference between continuous and discontinuous functions, and understand the properties of limits.

When a continuous variation of the argument asserts a constant variation in the value of the function, It is called a continuous function. Continuous functions have no sharp changes in value. In continuous functions, a small change in the argument produces a small change in its value. Discontinuous is a function that is not continuous.

When a function approaches a number it is called the limit. For example a function $f(x) = 4(x)$, and the limit of the function f(x) is $x$ approaches $3$ is $12$, symbolically, it is written as;

\[ \underset{x \rightarrow 3}{lim} f(x) = 4(3) = 12 \]

Expert Answer

Given that the function $f(x) = x + \sqrt{x-4}$ is defined on the interval $[4, \infty]$.

For $a > 4$ we have:

\[ \underset{x \rightarrow a}{lim} \space f(x) = \underset{x \rightarrow a}{lim} \space (x+ \sqrt{x-4}) \]

\[=\underset{x \rightarrow a}{lim} \space x+\underset{x \rightarrow a}{lim} \space (\sqrt{x-4}) \]

\[= \underset{x \rightarrow a}{lim} \space x+ \sqrt{\underset{x \rightarrow a}{lim} \space (x-4)} \]

\[=\underset{x \rightarrow a}{lim} \space x+ \sqrt{\underset{x \rightarrow a}{lim} \space x-\underset{x \rightarrow a}{lim} \space 4} \]

\[= a + \sqrt{a-4} \]

\[ f(a) \]

So the $\underset{x \rightarrow a}{lim} \space f(x) = f(a)$ for the all values of $a>4$. Therefore $f$ is continuous at $x=a$ for every $a$ in $(4, \infty)$.

Now checking at $\underset{x \rightarrow 4^+}{lim} \space f(x)$:

\[ \underset{x \rightarrow 4^+}{lim} \space f(x) = \underset{x \rightarrow 4^+}{lim} \space (x + \sqrt{x – 4}) \]

\[ = 4+\sqrt{4-4} \]

\[= 4+0\]

\[ = 4\]

\[= f(4)\]

So the $\underset{x \rightarrow 4^+}{lim} \space f(x) = 4$ Therefore, $f$ is continuous at 4$.

Numerical Answer

The function $f(x)= x+ \sqrt{x-4}$ is continuous at all points in the interval $[4, \infty]$. Therefore, $f$ is continuous at $x= a$ for every $a$ in $(4, \infty)$. Also, $\underset{x \rightarrow 4^+}{lim} \space f(x) = 4$ so the $f$ is continuous at $4$.

Thus, the function is continuous on $(4, \infty)$


Use the properties of limits and the definition of continuity to prove that the function $h(t)= \dfrac{2t-3t^2}{1+t^3}$ is continuous at the number $a=1$.

We have to show that for the function $h(t)= \dfrac{2t-3t^2}{1+t^3}$ we get $\underset{t \rightarrow 1}{lim} \space h(t) = h(1)$

\[ \underset{t \rightarrow 1}{lim} \space h(t) = \underset{t \rightarrow 1}{lim} \space \dfrac{2t – 3t^2}{1+t^3} \]

\[ \dfrac{\underset{t \rightarrow 1}{lim} \space (2t – 3t^2)} {\underset{t \rightarrow 1}{lim} \space (1+t^3) }\]

\[ \dfrac{2 \space \underset{t \rightarrow 1}{lim} \space (t) \space – 3 \space \underset{t \rightarrow 1}{lim} \space (t^2)} {\underset{t \rightarrow 1}{lim} \space (1)+ \space \underset{t \rightarrow 1}{lim} \space (t^3) }\]

\[ \dfrac{2 \space \underset{t \rightarrow 1}{lim} \space (t) \space – 3 \space (\underset{t \rightarrow 1}{lim} \space (t))^2} {\underset{t \rightarrow 1}{lim} \space (1)+ \space (\underset{t \rightarrow 1}{lim} \space (t) )^3}\]

\[= \dfrac{2(1)-3(1)^2}{(1) + (1)^3}\]

\[\underset{t \rightarrow 1}{lim} \space h(t)= \dfrac{2(1) – 3(1)^2}{(1) + (1)^3}=h(1)\]

Hence, proved that the function $h(t)= \dfrac{2t-3t^2}{1+t^3}$ is continuous at the number $a=1$.

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