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Let h(x)=(x)/(10sin^(2)(x+1)). Select the correct description of the one-sided limits of h at x=-1. Choose 1 answer: (A) lim_(x rarr-1^(+))h(x)=+oo and lim_(x rarr-1^(-))h(x)=+oo (B) lim_(x rarr-1^(+))h(x)=+oo and lim_(x rarr-1^(-))h(x)=-oo (C) lim_(x rarr-1^(+))h(x)=-oo and lim_(x rarr-1^(-))h(x)=+oo (D) lim_(x rarr-1^(+))h(x)=-oo and lim_(x rarr-1^(-))h(x)=-oo

Let h(x)=x10sin⁡2(x+1) h(x)=\frac{x}{10 \sin ^{2}(x+1)} .\newlineSelect the correct description of the one-sided limits of h h at x=−1 x=-1 .\newlineChoose 11 answer:\newline(A) lim⁡x→−1+h(x)=+∞ \lim _{x \rightarrow-1^{+}} h(x)=+\infty and lim⁡x→−1−h(x)=+∞ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \newline(B) lim⁡x→−1+h(x)=+∞ \lim _{x \rightarrow-1^{+}} h(x)=+\infty and lim⁡x→−1−h(x)=−∞ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \newline(C) lim⁡x→−1+h(x)=−∞ \lim _{x \rightarrow-1^{+}} h(x)=-\infty and lim⁡x→−1−h(x)=+∞ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \newline(D) lim⁡x→−1+h(x)=−∞ \lim _{x \rightarrow-1^{+}} h(x)=-\infty and lim⁡x→−1−h(x)=−∞ \lim _{x \rightarrow-1^{-}} h(x)=-\infty

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Q. Let h(x)=x10sin⁡2(x+1) h(x)=\frac{x}{10 \sin ^{2}(x+1)} .\newlineSelect the correct description of the one-sided limits of h h at x=−1 x=-1 .\newlineChoose 11 answer:\newline(A) lim⁡x→−1+h(x)=+∞ \lim _{x \rightarrow-1^{+}} h(x)=+\infty and lim⁡x→−1−h(x)=+∞ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \newline(B) lim⁡x→−1+h(x)=+∞ \lim _{x \rightarrow-1^{+}} h(x)=+\infty and lim⁡x→−1−h(x)=−∞ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \newline(C) lim⁡x→−1+h(x)=−∞ \lim _{x \rightarrow-1^{+}} h(x)=-\infty and lim⁡x→−1−h(x)=+∞ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \newline(D) lim⁡x→−1+h(x)=−∞ \lim _{x \rightarrow-1^{+}} h(x)=-\infty and lim⁡x→−1−h(x)=−∞ \lim _{x \rightarrow-1^{-}} h(x)=-\infty
  1. Given Function: h(x)=x10sin⁡2(x+1)h(x) = \frac{x}{10\sin^2(x+1)}. We need to find the limits as xx approaches −1-1 from the right (++) and from the left (−-).
  2. Approaching −1-1 from Right: For xx approaching −1-1 from the right (+), we look at values of xx that are just greater than −1-1. As xx gets close to −1-1, x+1x+1 gets close to 00. Since sin⁡(0)=0\sin(0) = 0, xx00 will also approach 00. The denominator of xx22 will approach 00, making xx22 grow without bound.
  3. Approaching −1-1 from Left: For xx approaching −1-1 from the left (−(-), we look at values of xx that are just less than −1-1. Similar to the previous step, as xx gets close to −1-1, x+1x+1 gets close to 00. Again, xx00 will approach 00, and the denominator of xx22 will approach 00, making xx22 grow without bound.
  4. Determining Sign of Infinity: We need to determine the sign of the infinity for the limits. Since xx is in the numerator, as xx approaches −1-1 from the right, xx is negative and getting closer to −1-1, so the numerator is negative. The denominator is always positive because it's sin⁡2(x+1)\sin^2(x+1). Therefore, the limit from the right is negative infinity.
  5. Final Limits: Similarly, as xx approaches −1-1 from the left, xx is still negative and getting closer to −1-1, so the numerator is negative. The denominator is positive because it's sin⁡2(x+1)\sin^2(x+1). Therefore, the limit from the left is also negative infinity.

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