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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)\newlinelim⁡x→−1+h(x)=+∞ and lim⁡x→−1−h(x)=+∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array} \newline(B)\newlinelim⁡x→−1+h(x)=+∞ and lim⁡x→−1−h(x)=−∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array} \newline(C)\newlinelim⁡x→−1+h(x)=−∞ and lim⁡x→−1−h(x)=+∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array} \newline(D)\newlinelim⁡x→−1+h(x)=−∞ and lim⁡x→−1−h(x)=−∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array}

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Full solution

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)\newlinelim⁡x→−1+h(x)=+∞ and lim⁡x→−1−h(x)=+∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array} \newline(B)\newlinelim⁡x→−1+h(x)=+∞ and lim⁡x→−1−h(x)=−∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array} \newline(C)\newlinelim⁡x→−1+h(x)=−∞ and lim⁡x→−1−h(x)=+∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array} \newline(D)\newlinelim⁡x→−1+h(x)=−∞ and lim⁡x→−1−h(x)=−∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array}
  1. Approach Analysis: To find the one-sided limits of h(x)h(x) as xx approaches −1-1, we need to consider the behavior of the function as xx approaches −1-1 from the left (x→−1−x \rightarrow -1^-) and from the right (x→−1+x \rightarrow -1^+).
  2. Limit Calculation: First, let's consider the limit as x→−1+x \rightarrow -1^+. We need to evaluate the sine function at the point x=−1x = -1. Since sin⁡(0)=0\sin(0) = 0, and the argument of the sine function is (x+1)(x+1), we have sin⁡(−1+1)=sin⁡(0)=0\sin(-1+1) = \sin(0) = 0.
  3. Limit from Right: As sin⁡(0)=0\sin(0) = 0, the denominator of h(x)h(x) becomes 10sin⁡2(x+1)=10sin⁡2(0)=010\sin^2(x+1) = 10\sin^2(0) = 0. Since the denominator approaches 00, the value of h(x)h(x) will approach infinity or negative infinity depending on the sign of the numerator.
  4. Limit from Left: The numerator is simply xx, which approaches −1-1 from the right. As xx approaches −1-1 from the right, the numerator is slightly greater than −1-1 and thus positive. Therefore, as the denominator approaches 00, the value of h(x)h(x) will approach positive infinity.
  5. Conclusion: Now, let's consider the limit as xx approaches −1-1 from the left (x→−1−x \rightarrow -1^-). The behavior of the sine function is the same since sin⁡(−1+1)=sin⁡(0)=0\sin(-1+1) = \sin(0) = 0. Therefore, the denominator will again approach 00.
  6. Conclusion: Now, let's consider the limit as xx approaches −1-1 from the left (x→−1−x \rightarrow -1^-). The behavior of the sine function is the same since sin⁡(−1+1)=sin⁡(0)=0\sin(-1+1) = \sin(0) = 0. Therefore, the denominator will again approach 00.As xx approaches −1-1 from the left, the numerator is slightly less than −1-1 and thus negative. However, since the denominator is squared, it will always be positive. Therefore, as the denominator approaches 00, the value of h(x)h(x) will approach negative infinity.
  7. Conclusion: Now, let's consider the limit as xx approaches −1-1 from the left (x→−1−x \rightarrow -1^-). The behavior of the sine function is the same since sin⁡(−1+1)=sin⁡(0)=0\sin(-1+1) = \sin(0) = 0. Therefore, the denominator will again approach 00.As xx approaches −1-1 from the left, the numerator is slightly less than −1-1 and thus negative. However, since the denominator is squared, it will always be positive. Therefore, as the denominator approaches 00, the value of h(x)h(x) will approach negative infinity.Based on the above analysis, we can conclude that the one-sided limits of h(x)h(x) as xx approaches −1-1 are as follows:\newline−1-133 and −1-144.

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