Respuesta :
Answer:
[tex]f(x)=\frac{x^2-25}{x+5}[/tex]
Step-by-step explanation:
Given:
We need to check the continuity of the functions at [tex]x=5[/tex].
Option 1:
[tex]f(x)=\frac{x^2-25}{x+5}[/tex]
Plug in 5 for 'x' and check the value of [tex]f(x)[/tex].
[tex]f(5)=\frac{5^2-25}{5+5}=\frac{25-25}{10}=0[/tex]
[tex]\lim_{x \to 5} \frac{x^2-25}{x+5}= \lim_{x \to 5} \frac{(x-5)(x+5)}{x+5}= \lim_{x \to 5} (x-5)=5-5=0[/tex]
So, the function is defined at and around [tex]x=5[/tex]and equal to 0. So, it is continuous at [tex]x=5[/tex].
Option 2:
[tex]f(x)=\left \{ {{\frac{x^2-25}{x-5}}\ x\ne 5 \atop {20}\ \ \ x=5} \right[/tex]
The given function is defined at [tex]x=5[/tex] and is equal to 20 but it has a different value around 5.
[tex]\lim_{x \to 5} \frac{x^2-25}{x-5}= \lim_{x \to 5} \frac{(x-5)(x+5)}{x-5}= \lim_{x \to 5} (x+5)=5+5=10[/tex]
So, the function has a value of 10 around [tex]x=5[/tex] and equal to 20 at [tex]x=5[/tex]. So, it is not continuous at [tex]x=5[/tex].
Option 3:
[tex]f(x)=\left \{ {{\frac{x^2-25}{x-5}}\ x\ne 5 \atop {0}\ \ \ x=5} \right[/tex]
The given function is defined at [tex]x=5[/tex] and is equal to 0 but it has a different value around 5.
[tex]\lim_{x \to 5} \frac{x^2-25}{x-5}= \lim_{x \to 5} \frac{(x-5)(x+5)}{x-5}= \lim_{x \to 5} (x+5)=5+5=10[/tex]
So, the function has a value of 10 around [tex]x=5[/tex] and equal to 0 at [tex]x=5[/tex]. So, it is not continuous at [tex]x=5[/tex].
