Limits Properties — Question 1

Question 1

Let the functions $f(x)$, $g(x)$, and $h(x)$ be defined by the following limits: $$\lim_{x \to 2} f(x) = 4, \quad \lim_{x \to 2} g(x) = -3, \quad \lim_{x \to 2} h(x) = 0$$

Evaluate the following limit expressions using limit laws and properties:

• $\displaystyle\lim_{x \to 2} \left[2f(x) + 5g(x)\right]$

• $\displaystyle\lim_{x \to 2} \left[\frac{f(x) \cdot g(x)}{h(x)}\right]$

• Explain whether the given information determines a signed infinite limit in part (b).

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

Question 1 - Solution

(a) Sum and constant-multiple laws

Using the given limits, $$\begin{aligned} \lim_{x\to2}\bigl[2f(x)+5g(x)\bigr] &=2\lim_{x\to2}f(x)+5\lim_{x\to2}g(x)\\ &=2(4)+5(-3)=\boxed{-7}. \end{aligned}$$

(b) Quotient with a zero denominator limit

The product law gives $$\lim_{x\to2}f(x)g(x)=4(-3)=-12, \qquad \lim_{x\to2}h(x)=0.$$

The quotient law does not apply because the denominator tends to zero. Since the numerator tends to a nonzero value, its magnitude stays bounded away from zero near $x=2$. Therefore, $$\left|\frac{f(x)g(x)}{h(x)}\right|\longrightarrow\infty.$$

$$\boxed{\text{The quotient has no finite limit.}}$$

(c) Is the sign of an infinite limit determined?

No. Take $f(x)=4$ and $g(x)=-3$. Each denominator below satisfies the given limit and is nonzero for $x\ne2$: $$\begin{array}{c|c} \text{Denominator }h(x)&\text{Behavior of }-12/h(x)\\[4pt]\hline (x-2)^2&\longrightarrow-\infty\\[4pt] -(x-2)^2&\longrightarrow+\infty\\[4pt] x-2&\text{opposite one-sided infinities} \end{array}$$

Thus more information about the sign of $h(x)$ is needed. $$\boxed{\text{A signed infinite limit is not determined.}}$$