Limits Properties — Question 3

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Question 3

Given the following limits: limx→1f(x)=4,limx→1g(x)=−2,limx→1h(x)=0\lim_{x \to 1} f(x) = 4, \quad \lim_{x \to 1} g(x) = -2, \quad \lim_{x \to 1} h(x) = 0

Define: k(x)=f(x)⋅h(x)+g(x)h(x)k(x) = \frac{f(x) \cdot h(x) + g(x)}{h(x)}

Evaluate lim⁡x→1k(x)\displaystyle \lim_{x \to 1} k(x), or explain why it does not exist.

Assume the quotient denominator is nonzero in some deleted neighborhood of the approach point. Distinguish finite limits from signed infinite limits.

Original worksheet page 1: question and worked solution for 2-4-003
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Question 3 - Solution

Step 1: Apply the product and sum laws

For the numerator, limx→1(f(x)h(x)+g(x))=4(0)+(−2)=−2.\lim_{x\to1}\bigl(f(x)h(x)+g(x)\bigr)=4(0)+(-2)=-2. The denominator satisfies lim⁡x→1h(x)=0\lim_{x\to1}h(x)=0.

Step 2: Determine whether a finite limit exists

The numerator tends to a nonzero number, so its magnitude stays bounded away from zero near x=1x=1. Since h(x)h(x) is nonzero in a deleted neighborhood, |k(x)|=|f(x)h(x)+g(x)h(x)|→∞.|k(x)|=\left|\frac{f(x)h(x)+g(x)}{h(x)}\right|\longrightarrow\infty. There is no finite limit.\boxed{\text{There is no finite limit.}}

Step 3: Examine the sign

The sign depends on the denominator. For example, choose f(x)=4f(x)=4 and g(x)=−2g(x)=-2, so that k(x)=4−2/h(x)k(x)=4-2/h(x). Denominator h(x)Behavior of k(x)(x−1)2→−∞−(x−1)2→+∞x−1opposite one-sided infinities\begin{array}{c|c} \text{Denominator }h(x)&\text{Behavior of }k(x)\\[4pt]\hline (x-1)^2&\longrightarrow-\infty\\[4pt] -(x-1)^2&\longrightarrow+\infty\\[4pt] x-1&\text{opposite one-sided infinities} \end{array}

All three choices satisfy the given limits. Therefore, A signed infinite limit is not determined.\boxed{\text{A signed infinite limit is not determined.}} Division by a zero limiting value is not a valid use of the quotient law.

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