Prove the continuity on an open interval

Question

I need to show, that function $f(x) =\frac{2x +3}{x-2}$ is continuous on the interval $(2,\infty)$

My attempt:

We should find the right-hand limit to prove the continuity: and this limit is equal to $\infty$. But in this case i don't now what to do next? Can you give me hints?

Answer 1

This function is a ratio. A ratio is continuous wherever its numerator and denominator are continuous and the denominator is not zero. (In symbols, $\frac{f(x)}{g(x)}$ is continuous at $x$ if $f$ and $g$ are continuous at $x$ and $g(x) \neq 0$. This is an application of the "quotient law" for limits to the ratio.) Your given numerator and denominator are polynomials, so are continuous for all values of $x$. Your denominator, $x-2$ is zero precisely when $x = 2$, so the ratio is continuous for all $x$ except $x=2$.

Written formally and specifically for the interval you give:

Let $x \in (2,\infty)$. Then using the product and sum laws of limits, $2x+3 = \lim_{t \rightarrow x} 2t+3$ and $x-2 = \lim_{t \rightarrow x} t-2$. Furthermore, since $x \neq 2$, the quotient law for limits gives $\frac{2x+3}{x-2} = \lim_{t \rightarrow x} \frac{2t+3}{t-2}$.

Answer 2

Let $I$ be an interval of $\mathbb{R}$ and $f,g \, : \, I \, \rightarrow \, \mathbb{R}$ two functions such that :

• $f$ and $g$ are continuous on $I$
• $\forall x \in I, \; g(x) \neq 0$

Then $f/g$ is continuous on $I$.

Take $f \, : \, x \, \mapsto \, 2x+3$ and $g \, : \, x \, \mapsto \, x-2$ and $I = (2,+\infty)$.

Answer 3

Use partial fractions to decompose the function as $f(x) = \frac{2(x - 2)}{x - 2} + \frac{7}{x - 2} = 2 + 7 . \frac{1}{x - 2}$. Since $\frac{1}{x}$ is continuous $\forall x > 0$, $\frac{1}{x - 2}$ will be continuous $\forall x > 2$.

Hence, $f(x), \ \forall x > 2$ is the sum of two continuous functions $g(x) = 2$ and $h(x) = \frac{1}{x - 2}$ and is hence continuous.

Answer 4

My suggestion is to show the points at which the function is discontinuous, and then show that these points are not contained in the open interval.

Answer 5

The function being a rational function will have values for all real $x$ except at $x=2$ since at $x=2$, the denominator becomes $0$. So find the left and right hand limits at $x=2$ to prove the continuity of the function. Nothing else is required.