Question

ii) Is the function continuous at the point? f(x) = \begin{cases} \frac{x^2 - 25}{x^2 + 8x + 15} & \text{if } x \neq -5 \\ x^3 + 120 & \text{if } x = -5 \end{cases} \lim_{x \to -5} f(x)

          ii) Is the function continuous at the point?

f(x) = \begin{cases} \frac{x^2 - 25}{x^2 + 8x + 15} & \text{if } x \neq -5 \\ x^3 + 120 & \text{if } x = -5 \end{cases}

\lim_{x \to -5} f(x)

ii) Is the function continuous at the point?

f(x) = (x^2 - 25)/(x^2 + 8x + 15) if x ≠-5
x^3 + 120 if x = -5

limx →-5 f(x)

Added by Richard A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Joseph David

Step 1

We can do this by plugging in x = 4 into the function: J(4) = 1(4) + 128 = 132 Show more…

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(x)J 1 x function '+128 continuous at the point? if if # # S- 4 Points

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Question

ii) Is the function continuous at the point? f(x) = \begin{cases} \frac{x^2 - 25}{x^2 + 8x + 15} & \text{if } x \neq -5 \\ x^3 + 120 & \text{if } x = -5 \end{cases} \lim_{x \to -5} f(x)

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