Question

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

• A. $\underset{x\to 5^{-}}{lim}f\left(x\right)=0$
• B. $\underset{x\to 5^{+}}{lim}f\left(x\right)=1$
• C. $\underset{x\to 5}{lim}f\left(x\right)$ does not exists
• D. $noneofthese$

Answer 1

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

The correct options are
A $\underset{x\to 5^{-}}{lim}f\left(x\right)=0$
B $\underset{x\to 5^{+}}{lim}f\left(x\right)=1$
C $\underset{x\to 5}{lim}f\left(x\right)$ does not exists

$\underset{x\to 5^{-}}{lim}f\left(x\right)$

$={lim}_{x\to 5^{-}}\frac{x^2-9x+20}{x-\left[x\right]}$

$=\underset{x\to 5^{-}}{lim}\frac{(x-5)(x-4)}{x-4}$

$=\underset{x\to 5^{-}}{lim}(x-5)=0$

Hence, option A is correct.

Now, $\underset{x\to 5^{-}}{lim}f\left(x\right)=0$

$\underset{x\to 5^{+}}{lim}\frac{x^2-9x+20}{x-\left[x\right]}$

$=\underset{x\to 5^{+}}{lim}\frac{(x-5)(x-4)}{x-5}$

$=\underset{x\to 5^{+}}{lim}(x-4)=1$

Hence, option B is correct.

Since, $LHL\ne RHL$

Hence, limit does not exist.