Question

which of the following limits are equal to - 1? i. $lim_{x
ightarrow0^{-}}\frac{|x|}{x}$ ii. $lim_{x
ightarrow3}\frac{x^{2}-7x + 12}{3 - x}$ iii. $lim_{x
ightarrowinfty}\frac{1 - x}{1 + x}$ (a) i only (b) i and iii only (c) ii and iii only (d) i, ii, iii

Explanation:

Step1: Analyze limit I

For $\lim_{x
ightarrow0^{-}}\frac{|x|}{x}$, when $x
ightarrow0^{-}$, $|x|=-x$. So $\lim_{x
ightarrow0^{-}}\frac{|x|}{x}=\lim_{x
ightarrow0^{-}}\frac{-x}{x}=- 1$.

Step2: Analyze limit II

For $\lim_{x
ightarrow3}\frac{x^{2}-7x + 12}{3 - x}=\lim_{x
ightarrow3}\frac{(x - 3)(x - 4)}{3 - x}=\lim_{x
ightarrow3}\frac{-(3 - x)(x - 4)}{3 - x}=\lim_{x
ightarrow3}-(x - 4)=1$.

Step3: Analyze limit III

For $\lim_{x
ightarrow\infty}\frac{1 - x}{1+x}$, divide both numerator and denominator by $x$: $\lim_{x
ightarrow\infty}\frac{\frac{1}{x}-1}{\frac{1}{x}+1}=\frac{0 - 1}{0 + 1}=-1$.

Answer:

B. I and III only