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5) $lim_{x ightarrow7}\frac{|7 - x|}{7 - x}$

Question

  1. $lim_{x

ightarrow7}\frac{|7 - x|}{7 - x}$

Explanation:

Step1: Analyze the absolute - value function

When \(x\to7\), consider two cases for \(|7 - x|\). If \(x\lt7\), then \(|7 - x|=7 - x\). If \(x\gt7\), then \(|7 - x|=-(7 - x)=x - 7\).

Step2: Calculate the left - hand limit (\(x\to7^{-}\))

When \(x\to7^{-}\), \(x\lt7\), so \(|7 - x| = 7 - x\). Then \(\lim_{x\to7^{-}}\frac{|7 - x|}{7 - x}=\lim_{x\to7^{-}}\frac{7 - x}{7 - x}=1\).

Step3: Calculate the right - hand limit (\(x\to7^{+}\))

When \(x\to7^{+}\), \(x\gt7\), so \(|7 - x|=x - 7\). Then \(\lim_{x\to7^{+}}\frac{|7 - x|}{7 - x}=\lim_{x\to7^{+}}\frac{x - 7}{7 - x}=- 1\).

Step4: Determine the limit

Since \(\lim_{x\to7^{-}}\frac{|7 - x|}{7 - x}
eq\lim_{x\to7^{+}}\frac{|7 - x|}{7 - x}\), the limit \(\lim_{x\to7}\frac{|7 - x|}{7 - x}\) does not exist.

Answer:

The limit does not exist.