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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)=1xtan(x) h(x)=\frac{1-x}{\tan (x)} .\newlineSelect the correct description of the one-sided limits of h h at x=0 x=0 .\newlineChoose 11 answer:\newline(A) limx0+h(x)=+ \lim _{x \rightarrow 0^{+}} h(x)=+\infty and limx0h(x)=+ \lim _{x \rightarrow 0^{-}} h(x)=+\infty \newline(B) limx0+h(x)=+ \lim _{x \rightarrow 0^{+}} h(x)=+\infty and limx0h(x)= \lim _{x \rightarrow 0^{-}} h(x)=-\infty \newline(C) limx0+h(x)= \lim _{x \rightarrow 0^{+}} h(x)=-\infty and limx0h(x)=+ \lim _{x \rightarrow 0^{-}} h(x)=+\infty \newline(D) limx0+h(x)= \lim _{x \rightarrow 0^{+}} h(x)=-\infty and limx0h(x)= \lim _{x \rightarrow 0^{-}} h(x)=-\infty

Full solution

Q. Let h(x)=1xtan(x) h(x)=\frac{1-x}{\tan (x)} .\newlineSelect the correct description of the one-sided limits of h h at x=0 x=0 .\newlineChoose 11 answer:\newline(A) limx0+h(x)=+ \lim _{x \rightarrow 0^{+}} h(x)=+\infty and limx0h(x)=+ \lim _{x \rightarrow 0^{-}} h(x)=+\infty \newline(B) limx0+h(x)=+ \lim _{x \rightarrow 0^{+}} h(x)=+\infty and limx0h(x)= \lim _{x \rightarrow 0^{-}} h(x)=-\infty \newline(C) limx0+h(x)= \lim _{x \rightarrow 0^{+}} h(x)=-\infty and limx0h(x)=+ \lim _{x \rightarrow 0^{-}} h(x)=+\infty \newline(D) limx0+h(x)= \lim _{x \rightarrow 0^{+}} h(x)=-\infty and limx0h(x)= \lim _{x \rightarrow 0^{-}} h(x)=-\infty
  1. Analyze function behavior approaching 00 from positive side: Let's analyze the behavior of the function h(x)=1xtan(x)h(x) = \frac{1-x}{\tan(x)} as xx approaches 00 from the positive side (right-hand limit). As xx approaches 00 from the right, tan(x)\tan(x) approaches 00 and since tan(x)\tan(x) is positive in the first quadrant, the denominator of h(x)h(x) approaches 00 from the positive side. The numerator xx00 approaches xx11 as xx approaches 00. Therefore, as xx approaches 00 from the right, h(x)h(x) approaches xx11 divided by a very small positive number, which means h(x)h(x) approaches positive infinity.
  2. Approaching 00 from the right: Now let's analyze the behavior of the function h(x)h(x) as xx approaches 00 from the negative side (left-hand limit). As xx approaches 00 from the left, tan(x)\tan(x) approaches 00 and since tan(x)\tan(x) is negative in the fourth quadrant, the denominator of h(x)h(x) approaches 00 from the negative side. The numerator xx00 approaches xx11 as xx approaches 00. Therefore, as xx approaches 00 from the left, h(x)h(x) approaches xx11 divided by a very small negative number, which means h(x)h(x) approaches negative infinity.

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