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Question Number 135216 by mohammad17 last updated on 11/Mar/21

$${lim}_{x\to 3^{+}}\frac{{\left[x\right]}^2-9}{x-3}$$

Answered by Olaf last updated on 11/Mar/21

$$\underset{x\to 3^{+}}{\lim }\frac{{\left[x\right]}^2-9}{x-3}=\underset{h\to 0^{+}}{\lim }\frac{{[3+h]}^2-9}{3+h-3}=\underset{h\to 0^{+}}{\lim }\frac{0^{+}}{h}=+\mathrm{\infty }$$

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