One–Sided Limits — Question 1

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Question 1

Let the function ff be defined as:

f(x)={x2−4x−2,if x<2x+2,if x≥2f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2}, & \text{if } x < 2 \\ x + 2, & \text{if } x \geq 2 \end{cases}

Evaluate: limx→2−f(x)andlimx→2+f(x)\lim_{x \to 2^-} f(x) \quad \text{and} \quad \lim_{x \to 2^+} f(x)

State whether lim⁡x→2f(x)\lim_{x \to 2} f(x) exists.

Original worksheet page 1: question and worked solution for 2-3-001
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Question 1 - Solution

We analyze the left-hand and right-hand limits separately.

Left-hand limit:

For x<2x < 2, f(x)=x2−4x−2=(x−2)(x+2)x−2f(x) = \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}

For x≠2x \neq 2, the x−2x - 2 cancels: f(x)=x+2f(x) = x + 2

So: limx→2−f(x)=limx→2−(x+2)=4\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 2) = 4

Right-hand limit:

For x≥2x \geq 2, we use the second piece: f(x)=x+2⇒limx→2+f(x)=4f(x) = x + 2 \Rightarrow \lim_{x \to 2^+} f(x) = 4

Conclusion:

Since both one-sided limits exist and are equal: limx→2f(x)=4\lim_{x \to 2} f(x) = \boxed{4}

Original worksheet page 2: question and worked solution for 2-3-001

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