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h(x)={[5x," for "x < -2],[x^(3)-2," for "x >= -2]:} Find lim_(x rarr-2)h(x). Choose 1 answer: (A) -10 (B) -2 (C) 10

h(x)={5x,amp;for xlt;−2 x3−2,amp;for x≥−2h(x)=\begin{cases} 5x, &amp; \text{for } x &lt; -2 \ x^{3}-2, &amp; \text{for } x \geq -2 \end{cases}\newline Find lim⁡x→−2h(x)\lim_{x \to -2}h(x).\newline Choose 11 answer:\newline (A) −10-10\newline (B) −2-2\newline (C) 1010

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Full solution

Q. h(x)={5x,for x<−2 x3−2,for x≥−2h(x)=\begin{cases} 5x, & \text{for } x < -2 \ x^{3}-2, & \text{for } x \geq -2 \end{cases}\newline Find lim⁡x→−2h(x)\lim_{x \to -2}h(x).\newline Choose 11 answer:\newline (A) −10-10\newline (B) −2-2\newline (C) 1010
  1. Define h(x)h(x): We need to find the limit of h(x)h(x) as xx approaches −2-2. The function h(x)h(x) is defined piecewise:\newlineh(x)={5xamp;for xlt;−2, x3−2amp;for x≥−2h(x) = \begin{cases} 5x &amp; \text{for } x &lt; -2,\ x^3 - 2 &amp; \text{for } x \geq -2 \end{cases}\newlineTo find the limit as xx approaches −2-2, we need to consider the value from both sides of −2-2.
  2. Limit from left side: First, let's find the limit from the left side xx approaching −2-2 from values less than −2-2. For x < -2, h(x)=5xh(x) = 5x.lim⁡x→−2−h(x)=lim⁡x→−2−5x=5×(−2)=−10\lim_{x \to -2^-} h(x) = \lim_{x \to -2^-} 5x = 5 \times (-2) = -10.
  3. Limit from right side: Now, let's find the limit from the right side xx approaching −2-2 from values greater than or equal to −2-2. For x≥−2x \geq -2, h(x)=x3−2h(x) = x^3 - 2.lim⁡x→−2+h(x)=lim⁡x→−2+(x3−2)=(−2)3−2=−8−2=−10\lim_{x \to -2^+} h(x) = \lim_{x \to -2^+} (x^3 - 2) = (-2)^3 - 2 = -8 - 2 = -10.
  4. Final limit: Since the limit from the left side and the limit from the right side are equal, the limit of h(x)h(x) as xx approaches −2-2 exists and is equal to −10-10.\newlinelim⁡x→−2h(x)=−10\lim_{x \to -2} h(x) = -10.

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