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

Use The Definition Of Continuity And The Properties Of Limits To Show That The Function

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

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 $a=1$.

Previous Question < > Next Question

5/5 - (6 votes)