Question

Let $f(x)=\frac{x^2-9x+20}{x-[x]}($ where $[x]$ is the greatest integer not greater than $x$ ), then

• A. $\underset{x\to 5^{-}}{\lim }f(x)=0$
• B. $\underset{x\to 5^{+}}{\lim }f(x)=1$
• C. $\underset{x\to 5}{\lim }f(x)$ does not exist
• D. None of these

Answer 1

Answer: (A), (B), (C)

$=\underset{x\to 5^{-}}{\lim }\frac{x^2-9x+20}{x-[x]}$
$=\underset{x\to 5^{-}}{\lim }\frac{(x-5)(x-4)}{x-4}$
$=\underset{x\to 5^{-}}{\lim }(x-5)=0$
$=\underset{x\to 5^{-}}{\lim }\frac{x^2-9x+20}{x-[x]}$
$=\underset{x\to 5^{+}}{\lim }\frac{(x-5)(x-4)}{x-5}$
$=\underset{x\to 5^{+}}{\lim }(x-4)=1$