Answer:
The function is continuous at a = 5
Explanation:
Given:
[tex]18)\text{ }f(x)\text{ = }\frac{2x^2+3x+1}{x^2+5x};\text{ a = 5}[/tex]
To find:
If the function is continuous at a = 5
For a function to be continuous at a point, the limit exists for the point and the value of the function at that point must be equal to the limit at the point.
when x = 5
[tex]\begin{gathered} f(x)\text{ = }\frac{2(5)^2+3(5)+1}{(5)^2+5(5)} \\ \\ f(x)\text{ = }\frac{50\text{ + 15 + 1}}{25\text{ + 25}} \\ \\ f(x)\text{ = }\frac{66}{50} \\ \\ f(x)\text{ = }\frac{33}{25} \end{gathered}[/tex]
Finding the limit at the point:
[tex]\begin{gathered} \lim_{a\to5}\frac{2x^2+3x\text{ + 1}}{x^2+5x} \\ \\ To\text{ get the limit at the point a = 5, we will susbtitute x with 5} \\ =\text{ }\frac{2(5)\placeholder{⬚}^2+3(5)+1}{(5)\placeholder{⬚}^2+5(5)} \\ \\ =\text{ }\frac{50+15+1}{25+25}\text{ = }\frac{66}{50} \\ \\ =\text{ }\frac{33}{25} \end{gathered}[/tex]
The value of the function at that point is equal to the limit at the point.
Hence, the function is continuous at a = 5
