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determine the following limits, using \\(\\infty\\) or \\(-\\infty\\) w…

Question

determine the following limits, using \\(\infty\\) or \\(-\infty\\) when appropriate, or state that they do not exist.
a. \\(\lim_{x \to 1^+} \frac{x^2 - 8x + 15}{(x - 1)^2}\\)
b. \\(\lim_{x \to 1^-} \frac{x^2 - 8x + 15}{(x - 1)^2}\\)
c. \\(\lim_{x \to 1} \frac{x^2 - 8x + 15}{(x - 1)^2}\\)

a. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. \\(\lim_{x \to 1^+} \frac{x^2 - 8x + 15}{(x - 1)^2} = \\) (simplify your answer.)
b. the limit does not exist and is neither \\(\infty\\) nor \\(-\infty\\).

b. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. \\(\lim_{x \to 1^-} \frac{x^2 - 8x + 15}{(x - 1)^2} = \\) (simplify your answer.)
b. the limit does not exist and is neither \\(\infty\\) nor \\(-\infty\\).

c. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. \\(\lim_{x \to 1} \frac{x^2 - 8x + 15}{(x - 1)^2} = \\) (simplify your answer.)
b. the limit does not exist and is neither \\(\infty\\) nor \\(-\infty\\).

Explanation:

Evaluate the limit of the numerator and denominator as \(x \to 1\)

$$ \lim_{x \to 1} (x^2 - 8x + 15) = 1^2 - 8(1) + 15 = 8 > 0 $$
$$ \lim_{x \to 1} (x - 1)^2 = 0 $$

Analyze the sign of the function near \(x = 1\)

For all \(x
eq 1\), \((x - 1)^2 > 0\).
Since the numerator is positive (\(8 > 0\)) and the denominator is positive and approaches \(0\) as \(x \to 1^+\) and \(x \to 1^-\):

$$ \lim_{x \to 1^+} \frac{x^2 - 8x + 15}{(x - 1)^2} = \infty $$
$$ \lim_{x \to 1^-} \frac{x^2 - 8x + 15}{(x - 1)^2} = \infty $$

Determine the two-sided limit

Since both one-sided limits are \(\infty\):

$$ \lim_{x \to 1} \frac{x^2 - 8x + 15}{(x - 1)^2} = \infty $$

Answer:

Question a

  • A. \(\lim_{x \to 1^+} \frac{x^2 - 8x + 15}{(x - 1)^2} = \infty\) (Correct answer)
  • B. The limit does not exist and is neither \(\infty\) nor \(-\infty\).

Question b

  • A. \(\lim_{x \to 1^-} \frac{x^2 - 8x + 15}{(x - 1)^2} = \infty\) (Correct answer)
  • B. The limit does not exist and is neither \(\infty\) nor \(-\infty\).

Question c

  • A. \(\lim_{x \to 1} \frac{x^2 - 8x + 15}{(x - 1)^2} = \infty\) (Correct answer)
  • B. The limit does not exist and is neither \(\infty\) nor \(-\infty\).