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Find limx5((x+1)28x) \lim_{x \to -5}\left(\frac{(x+1)^{2}}{8-x}\right) .\newlineChoose 11 answer:\newline(A) 1613 \frac{16}{13} \newline(B) 2 2 \newline(C) 12 12 \newline(D) The limit doesn't exist

Full solution

Q. Find limx5((x+1)28x) \lim_{x \to -5}\left(\frac{(x+1)^{2}}{8-x}\right) .\newlineChoose 11 answer:\newline(A) 1613 \frac{16}{13} \newline(B) 2 2 \newline(C) 12 12 \newline(D) The limit doesn't exist
  1. Check Validity of Substitution: To find the limit of the function (x+1)2/(8x)(x+1)^2/(8-x) as xx approaches 5-5, we can directly substitute the value of xx with 5-5, as long as the substitution does not result in a division by zero or an indeterminate form.
  2. Substitute xx with 5-5: Substitute xx with 5-5 into the function (x+1)28x\frac{(x+1)^2}{8-x}.((5)+1)28(5)=(4)28+5=1613\frac{((-5)+1)^2}{8-(-5)} = \frac{(4)^2}{8+5} = \frac{16}{13}.
  3. Verify Valid Numerical Value: Since the substitution resulted in a valid numerical value and did not cause any indeterminate form or division by zero, the limit exists and is equal to the calculated value.
  4. Final Answer: The final answer is 1613\frac{16}{13}, which corresponds to option (A).

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