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Let h(x)=-(2x)/((x+1)^(2)). 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)=−2x(x+1)2 h(x)=-\frac{2 x}{(x+1)^{2}} .\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)=−2x(x+1)2 h(x)=-\frac{2 x}{(x+1)^{2}} .\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. Analyze Behavior as xx Approaches −1-1: Analyze the behavior of h(x)h(x) as xx approaches −1-1 from the right (x→−1+x \to -1^+).\newlineWe need to consider the sign of the numerator and the denominator separately as xx approaches −1-1 from the right. The numerator, −2x-2x, will be negative since xx is close to −1-1. The denominator, −1-111, will be a small positive number squared, which is also positive. A negative divided by a positive is negative, so the limit will approach negative infinity.
  2. Calculate Right-Hand Limit: Calculate the right-hand limit of h(x)h(x) as xx approaches −1-1.
    lim⁡x→−1+h(x)=lim⁡x→−1+−2x(x+1)2\lim_{x \to -1^+} h(x) = \lim_{x \to -1^+} -\frac{2x}{(x+1)^2}
    As xx approaches −1-1 from the right, the denominator approaches 00 and the expression becomes a negative number divided by a positive number that is approaching 00, which tends to negative infinity.
    lim⁡x→−1+h(x)=−∞\lim_{x \to -1^+} h(x) = -\infty
  3. Analyze Behavior as xx Approaches −1-1: Analyze the behavior of h(x)h(x) as xx approaches −1-1 from the left (x→−1−x \to -1^-).\newlineWe need to consider the sign of the numerator and the denominator separately as xx approaches −1-1 from the left. The numerator, −2x-2x, will be negative since xx is close to −1-1. The denominator, −1-111, will again be a small positive number squared, which is also positive. A negative divided by a positive is negative, so the limit will approach negative infinity.
  4. Calculate Left-Hand Limit: Calculate the left-hand limit of h(x)h(x) as xx approaches −1-1.
    lim⁡x→−1−h(x)=lim⁡x→−1−−2x(x+1)2\lim_{x \to -1^-} h(x) = \lim_{x \to -1^-} -\frac{2x}{(x+1)^2}
    As xx approaches −1-1 from the left, the denominator approaches 00 and the expression becomes a negative number divided by a positive number that is approaching 00, which tends to negative infinity.
    lim⁡x→−1−h(x)=−∞\lim_{x \to -1^-} h(x) = -\infty

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