Question

Solve:

${lim}_{x\to 4^{+}}\frac{x^2-7x+12}{x-\left[x\right]}=$ [where $[.]$ denotes G.I.F]

• A. $1$
• B. $0$
• C. $2$
• D. does not exist

Answer 1

The correct option is A $1$

$L=\underset{x\to 4^{+}}{lim}\frac{x^2-7x+12}{x-\left[x\right]}$

$=\underset{x\to 4^{+}}{lim}\frac{x^2-7x+12}{x-4}$

$=\underset{x\to 4^{+}}{lim}\frac{(x-3)(x-4)}{(x-4)}$

$=\underset{x\to 4^{+}}{lim}x-3$

$=4-3$

$=1$

$\therefore L=1$