Find $$ \delta $$ corresponding to $$ \epsilon $$ for a proof that $$ \lim _{x \rightarrow 11^{-}} \sqrt{11-x}=0 $$ $$ \delta=\frac{\epsilon}{11} $$ $$ \delta=11-\varepsilon $$ $$ \delta=\epsilon $$ $$ \delta=11 \epsilon $$ $$ \delta=\epsilon^{2} $$

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To prove that $$ \lim_{x \rightarrow 11^{-}} \sqrt{11-x} = 0, $$ we must show that for every $$\epsilon > 0$$, there exists a $$\delta > 0$$ such that if $$ 0 < 11-x < \delta, $$ then $$ \sqrt{11-x} < \epsilon. $$ Starting with the inequality $$ \sqrt{11-x} < \epsilon, $$ square both sides (the square function is increasing for positive numbers) to obtain $$ 11-x < \epsilon^2. $$ This implies that if we choose $$ \delta = \epsilon^2, $$ then whenever $$0 < 11-x < \delta$$, it follows that $$ \sqrt{11-x} < \epsilon. $$ Thus, the correct choice corresponding to $$\epsilon$$ is $$ \delta = \epsilon^2. $$ $$ \delta = \epsilon^{2} $$

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