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Let 
f(x)=-(4)/(x-1).
Select the correct description of the one-sided limits of 
f at 
x=1.
Choose 1 answer:
(A) 
lim_(x rarr1^(+))f(x)=+oo and 
lim_(x rarr1^(-))f(x)=+oo
(B) 
lim_(x rarr1^(+))f(x)=+oo and 
lim_(x rarr1^(-))f(x)=-oo
(C) 
lim_(x rarr1^(+))f(x)=-oo and 
lim_(x rarr1^(-))f(x)=+oo
(D) 
lim_(x rarr1^(+))f(x)=-oo and 
lim_(x rarr1^(-))f(x)=-oo

Let f(x)=4x1 f(x)=-\frac{4}{x-1} .\newlineSelect the correct description of the one-sided limits of f f at x=1 x=1 .\newlineChoose 11 answer:\newline(A) limx1+f(x)=+ \lim _{x \rightarrow 1^{+}} f(x)=+\infty and limx1f(x)=+ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \newline(B) limx1+f(x)=+ \lim _{x \rightarrow 1^{+}} f(x)=+\infty and limx1f(x)= \lim _{x \rightarrow 1^{-}} f(x)=-\infty \newline(C) limx1+f(x)= \lim _{x \rightarrow 1^{+}} f(x)=-\infty and limx1f(x)=+ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \newline(D) limx1+f(x)= \lim _{x \rightarrow 1^{+}} f(x)=-\infty and limx1f(x)= \lim _{x \rightarrow 1^{-}} f(x)=-\infty

Full solution

Q. Let f(x)=4x1 f(x)=-\frac{4}{x-1} .\newlineSelect the correct description of the one-sided limits of f f at x=1 x=1 .\newlineChoose 11 answer:\newline(A) limx1+f(x)=+ \lim _{x \rightarrow 1^{+}} f(x)=+\infty and limx1f(x)=+ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \newline(B) limx1+f(x)=+ \lim _{x \rightarrow 1^{+}} f(x)=+\infty and limx1f(x)= \lim _{x \rightarrow 1^{-}} f(x)=-\infty \newline(C) limx1+f(x)= \lim _{x \rightarrow 1^{+}} f(x)=-\infty and limx1f(x)=+ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \newline(D) limx1+f(x)= \lim _{x \rightarrow 1^{+}} f(x)=-\infty and limx1f(x)= \lim _{x \rightarrow 1^{-}} f(x)=-\infty
  1. Analyze Function: Analyze the function near the point of interest.\newlineWe have the function f(x)=4x1f(x) = -\frac{4}{x-1}. We need to find the one-sided limits as xx approaches 11 from the left (x1x \to 1^-) and from the right (x1+x \to 1^+).
  2. Limit Right Approach: Consider the limit as xx approaches 11 from the right (x1+x \to 1^+).\newlineAs xx gets closer to 11 from the right, the denominator (x1x-1) becomes a small positive number, and since the numerator is 4-4, the fraction becomes a large negative number. Therefore, the limit from the right is negative infinity.\newlinelimx1+f(x)=\lim_{x \to 1^+} f(x) = -\infty
  3. Limit Left Approach: Consider the limit as xx approaches 11 from the left (x1x \to 1^-).\newlineAs xx gets closer to 11 from the left, the denominator (x1x-1) becomes a small negative number, and since the numerator is 4-4, the fraction becomes a large positive number. Therefore, the limit from the left is positive infinity.\newlinelimx1f(x)=+\lim_{x \to 1^-} f(x) = +\infty

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