Let h(x)=(1-x)/(tan(x)). Select the correct description of the one-sided limits of h at x=0. Choose 1 answer: (A) lim_(x rarr0^(+))h(x)=+oo and lim_(x rarr0^(-))h(x)=+oo (B) lim_(x rarr0^(+))h(x)=+oo and lim_(x rarr0^(-))h(x)=-oo (C) lim_(x rarr0^(+))h(x)=-oo and lim_(x rarr0^(-))h(x)=+oo (D) lim_(x rarr0^(+))h(x)=-oo and lim_(x rarr0^(-))h(x)=-oo

Analyze function behavior approaching 0 from positive side: Let's analyze the behavior of the function h(x) = (1-x)/tan(x) as x approaches 0 from the positive side (right-hand limit). As x approaches 0 from the right, tan(x) approaches 0 and since tan(x) is positive in the first quadrant, the denominator of h(x) approaches 0 from the positive side. The numerator (1-x) approaches 1 as x approaches 0. Therefore, as x approaches 0 from the right, h(x) approaches 1 divided by a very small positive number, which means h(x) approaches positive infinity.Approaching 0 from the right: Now let's analyze the behavior of the function h(x) as x approaches 0 from the negative side (left-hand limit). As x approaches 0 from the left, tan(x) approaches 0 and since tan(x) is negative in the fourth quadrant, the denominator of h(x) approaches 0 from the negative side. The numerator (1-x) approaches 1 as x approaches 0. Therefore, as x approaches 0 from the left, h(x) approaches 1 divided by a very small negative number, which means h(x) approaches negative infinity.

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Let
h(x)=(1-x)/(tan(x)).
Select the correct description of the one-sided limits of
h at
x=0.
Choose 1 answer:
(A)
lim_(x rarr0^(+))h(x)=+oo and
lim_(x rarr0^(-))h(x)=+oo
(B)
lim_(x rarr0^(+))h(x)=+oo and
lim_(x rarr0^(-))h(x)=-oo
(C)
lim_(x rarr0^(+))h(x)=-oo and
lim_(x rarr0^(-))h(x)=+oo
(D)
lim_(x rarr0^(+))h(x)=-oo and
lim_(x rarr0^(-))h(x)=-oo

Let $h(x)=\frac{1-x}{\tan (x)}$ . $\newline$ Select the correct description of the one-sided limits of $h$ at $x=0$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\lim _{x \rightarrow 0^{+}} h(x)=+\infty$ and $\lim _{x \rightarrow 0^{-}} h(x)=+\infty$ $\newline$ (B) $\lim _{x \rightarrow 0^{+}} h(x)=+\infty$ and $\lim _{x \rightarrow 0^{-}} h(x)=-\infty$ $\newline$ (C) $\lim _{x \rightarrow 0^{+}} h(x)=-\infty$ and $\lim _{x \rightarrow 0^{-}} h(x)=+\infty$ $\newline$ (D) $\lim _{x \rightarrow 0^{+}} h(x)=-\infty$ and $\lim _{x \rightarrow 0^{-}} h(x)=-\infty$

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Math Problems

Algebra 1

Compare linear, exponential, and quadratic growth

Full solution

Q. Let

h(x)=\frac{1-x}{\tan (x)}

\newline

Select the correct description of the one-sided limits of

h

x=0

\newline

Choose

1

answer:

\newline

(A)

\lim _{x \rightarrow 0^{+}} h(x)=+\infty

and

\lim _{x \rightarrow 0^{-}} h(x)=+\infty

\newline

(B)

\lim _{x \rightarrow 0^{+}} h(x)=+\infty

and

\lim _{x \rightarrow 0^{-}} h(x)=-\infty

\newline

(C)

\lim _{x \rightarrow 0^{+}} h(x)=-\infty

and

\lim _{x \rightarrow 0^{-}} h(x)=+\infty

\newline

(D)

\lim _{x \rightarrow 0^{+}} h(x)=-\infty

and

\lim _{x \rightarrow 0^{-}} h(x)=-\infty

Analyze function behavior approaching $0$ from positive side: Let's analyze the behavior of the function $h(x) = \frac{1-x}{\tan(x)}$ as $x$ approaches $0$ from the positive side (right-hand limit). As $x$ approaches $0$ from the right, $\tan(x)$ approaches $0$ and since $\tan(x)$ is positive in the first quadrant, the denominator of $h(x)$ approaches $0$ from the positive side. The numerator $x$ $0$ approaches $x$ $1$ as $x$ approaches $0$ . Therefore, as $x$ approaches $0$ from the right, $h(x)$ approaches $x$ $1$ divided by a very small positive number, which means $h(x)$ approaches positive infinity.
Approaching $0$ from the right: Now let's analyze the behavior of the function $h(x)$ as $x$ approaches $0$ from the negative side (left-hand limit). As $x$ approaches $0$ from the left, $\tan(x)$ approaches $0$ and since $\tan(x)$ is negative in the fourth quadrant, the denominator of $h(x)$ approaches $0$ from the negative side. The numerator $x$ $0$ approaches $x$ $1$ as $x$ approaches $0$ . Therefore, as $x$ approaches $0$ from the left, $h(x)$ approaches $x$ $1$ divided by a very small negative number, which means $h(x)$ approaches negative infinity.