Limits Properties — Question 3

Question 3

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

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

Evaluate $\displaystyle \lim_{x \to 1} k(x)$, 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.

Question 3 - Solution

Step 1: Apply the product and sum laws

For the numerator, $$\lim_{x\to1}\bigl(f(x)h(x)+g(x)\bigr)=4(0)+(-2)=-2.$$ The denominator satisfies $\lim_{x\to1}h(x)=0$.

Step 2: Determine whether a finite limit exists

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

Step 3: Examine the sign

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

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