Правый предел с модулем при 7

Problem

Evaluate

$$\lim_{x \to 7^+}\left(1+\frac{|x-7|}{7-x}\right).$$

Step 1: Use the Approach Direction

Since $x \to 7^+$, $x$ approaches $7$ from the right. That means $x>7$ while it gets closer and closer to $7$.

Therefore,

$$x-7>0.$$

Step 2: Simplify the Absolute Value

Because $x-7$ is positive, the absolute value does not change its sign:

$$|x-7|=x-7.$$

So the expression becomes

$$1+\frac{x-7}{7-x}.$$

Step 3: Rewrite the Denominator

Compare the numerator and denominator:

$$7-x=-(x-7).$$

Therefore,

$$\frac{x-7}{7-x} = \frac{x-7}{-(x-7)} = -1.$$

Step 4: Combine the Terms

So the expression simplifies to

$$1+(-1)=0.$$

Therefore,

$$\lim_{x \to 7^+}\left(1+\frac{|x-7|}{7-x}\right)=0.$$