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(A) We need to evaluate $\lim _{x \rightarrow 3} \frac{11-[2-x]}{[x+10]}$.First,consider the left-hand limit $(LHL)$ as $x \rightarrow 3^-$:As $x \rig

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(A) We need to evaluate $\lim _{x \rightarrow 3} \frac{11-[2-x]}{[x+10]}$.First,consider the left-hand limit $(LHL)$ as $x \rightarrow 3^-$:As $x \rightarrow 3^-$,$x$ is slightly less than $3$,so $2-x$ is slightly greater than $-1$. Thus,$[2-x] = -1$.Also,$x+10$ is slightly less than $13$,so $[x+10] = 12$.$LHL = \lim _{x \rightarrow 3^-} \frac{11-(-1)}{12} = \frac{12}{12} = 1$.Now,consider the right-hand limit $(RHL)$ as $x \rightarrow 3^+$:As $x \rightarrow 3^+$,$x$ is slightly greater than $3$,so $2-x$ is slightly less than $-1$. Thus,$[2-x] = -2$.Also,$x+10$ is slightly greater than $13$,so $[x+10] = 13$.$RHL = \lim _{x \rightarrow 3^+} \frac{11-(-2)}{13} = \frac{13}{13} = 1$.Since $LHL = RHL = 1$,the limit exists and is equal to $1$.

If $[t]$ represents the greatest integer $\leq t$,then the value of $\lim _{x \rightarrow 3} \frac{11-[2-x]}{[x+10]}$ is

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