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MTH 254 Calculus IV Winter 2026

Richard Zhao

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Worksheet Assigned Problems for Section 14.2

The problems listed below are assigned to be included in your problem set portfolio. Note that a specific selection of these problems will also form the written homework assignments. I recommend working through all of them to ensure a solid grasp of the material. Reach out to Richard for help if you get stuck or have any questions.
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The solutions will be posted after the written homework due dates. If you have any questions about your work, talk to Richard and he is happy to discuss the process with you.
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14.2.7.

Evaluate the following limit using continuity
\begin{equation*} \lim_{(x,y) \to (1,1)} \frac{e^{x^2} - e^{-y^2}}{x + y} \end{equation*}
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Solution.
The function is the quotient of two continuous functions, and the denominator is not zero at the point \((1,1)\text{.}\) Therefore, the function is continuous at this point, and we may compute the limit by substitution.
\begin{equation*} \lim_{(x,y) \to (1,1)} \frac{e^{x^2} - e^{-y^2}}{x + y} = \frac{e^{1^2} - e^{-1^2}}{1 + 1} = \frac{e - e^{-1}}{2} \end{equation*}
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14.2.13.

Does \(\ds \lim_{(x,y) \to (0,0)} \dfrac{y^2}{x^2 + y^2}\) exist? Explain.
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Solution.
This limit does not exist. Consider the following paths to the point \((x,y) = (0,0)\text{.}\) First along the line \(x = 0\) and second along the line \(y = x\text{.}\)
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First along the line \(x = 0\text{,}\) we have
\begin{equation*} \lim_{y \to 0} \dfrac{y^2}{0^2 + y^2} = \lim_{y \to 0} 1 = 1 \end{equation*}
So along the line \(x = 0\text{,}\) the limit is 1.
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Next along the line \(y = x\text{,}\) we have
\begin{equation*} \lim_{x \to 0} \dfrac{x^2}{x^2 + x^2} = \lim_{x \to 0} \dfrac{1}{2} = \dfrac{1}{2} \end{equation*}
So along the line \(y = x\text{,}\) the limit is \(\dfrac{1}{2}\text{.}\)
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Since these two limits are not equal, the limit in question does not exist.
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14.2.15.

Let \(f(x,y) = \dfrac{x^3 + y^3}{xy^2}\text{.}\) Set \(y = mx\) and show that the resulting limit depends on \(m\text{,}\) and therefore the limit \(\ds \lim_{(x,y) \to (0,0)} f(x,y)\) does not exist.
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Solution.
We have
\begin{equation*} f(x,mx) = \frac{x^3 + (mx)^3}{x(mx)^2} = \frac{\lp m^3 + 1 \rp x^3}{m^2x^3} = \frac{m^3 + 1}{m^2} \end{equation*}
So \(f\) is a constant along any line through the origin. Since \(\dfrac{m^3 + 1}{m^2}\) is not a constant function of \(m\text{,}\) the limit of \(f(x,mx)\) depends on \(m\text{.}\) For example,
\begin{align*} \text{When } m = 2, \, \amp \lim_{x\to 0} f(x.mx) = \lim_{x\to 0} \frac{2^3 + 1}{2^2} = \frac{9}{4} \\ \text{When } m = 1, \, \amp \lim_{x\to 0} f(x.mx) = \lim_{x\to 0} \frac{1^3 + 1}{1^2} = 2 \end{align*}
Therefore, the limit in question does not exist.
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14.2.21.

Evaluate the limit \(\ds \lim_{(x,y) \to (0,0)} \dfrac{xy}{3x^2 + 2y^2}\) or show that it does not exist.
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Solution.
This limit does not exist. Consider the limit as \((x,y) \to (0,0)\) along the line \(y = mx\text{.}\)
\begin{equation*} \lim_{x\to 0} \frac{x(mx)}{3x^2 + 2(mx)^2} = \lim_{x\to 0} \frac{mx^2}{x^2 \lp 3 + 2m^2 \rp} = \lim_{x\to 0}\frac{m}{3+2m^2} \end{equation*}
This value is different for different values of \(m\text{,}\) so the limit depends on the path taken and the limit in question does not exist.
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14.2.25.

Use the Squeeze Theorem to evaluate
\begin{equation*} \lim_{(x,y) \to (4,0)} \lp x^2 - 16 \rp \cos\lp \frac{1}{(x - 4)^2 + y^2} \rp \end{equation*}
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Solution.
Consider the following inequalities:
\begin{equation*} -1 \leq \cos\lp \frac{1}{(x - 4)^2 + y^2} \rp \leq 1 \end{equation*}
Then for \(x\) such that \(x \geq 4\) then \(x^2 - 16 \geq 0\text{,}\) so we have
\begin{gather*} -\lp x^2 - 16 \rp \leq \lp x^2 - 16 \rp \cos\lp \frac{1}{(x - 4)^2 + y^2} \rp \leq \lp x^2 - 16 \rp \\ \lim_{(x,y) \to (4,0)} -\lp x^2 - 16 \rp \leq \lim_{(x,y) \to (4,0)} \lp x^2 - 16 \rp \cos\lp \frac{1}{(x - 4)^2 + y^2} \rp \leq \lim_{(x,y) \to (4,0)} \lp x^2 - 16 \rp \end{gather*}
Then the two limits at the ends of the inequality are clearly equal to \(0\text{,}\) by the Squeeze Theorem.
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Now, if \(x \lt 4\text{,}\) then \(x^2 - 16 \lt 0\text{,}\) so we have
\begin{gather*} \lp x^2 - 16 \rp \leq \lp x^2 - 16 \rp \cos\lp \frac{1}{(x - 4)^2 + y^2} \rp \leq -\lp x^2 - 16 \rp \\ \lim_{(x,y) \to (4,0)} \lp x^2 - 16 \rp \leq \lim_{(x,y) \to (4,0)} \lp x^2 - 16 \rp \cos\lp \frac{1}{(x - 4)^2 + y^2} \rp \leq \lim_{(x,y) \to (4,0)} -\lp x^2 - 16 \rp \end{gather*}
Then the two limits at the ends of the inequality are clearly equal to \(0\text{,}\) by the Squeeze Theorem.
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Thus, we can conclude
\begin{equation*} \lim_{(x,y) \to (4,0)} \lp x^2 - 16 \rp \cos\lp \frac{1}{(x - 4)^2 + y^2} \rp = 0 \end{equation*}
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Exercise Group.

For the following exercises, evaluate the limit or determine that it does not exist.
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14.2.31.
\(\ds \lim_{(x,y) \to (3,4)} \dfrac{1}{\sqrt{x^2 + y^2}}\)
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Solution.
The function \(\dfrac{1}{\sqrt{x^2 + y^2}}\) is continuous at the point \((3,4)\) since it is the quotient of two continuous functions and the denominator is not zero at \((3,4)\text{.}\) We compute the limit by substitution:
\begin{equation*} \lim_{(x,y) \to (3,4)} \dfrac{1}{\sqrt{x^2 + y^2}} = \dfrac{1}{\sqrt{3^2 + 4^2}} = \dfrac{1}{5} \end{equation*}
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14.2.33.
\(\ds \lim_{(x,y) \to (\pi,0)} \dfrac{\cos(x)}{\sin(y)}\)
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Solution.
This limit does not exist. As \((x,y)\) approaches \((\pi,0)\text{,}\) the numerator approaches \(\cos\lp \pi \rp = -1\) and the denominator approaches \(\sin\lp 0 \rp = 0\text{.}\) This form \(-1/0\) indicates the limit will be unbounded. We can investigate the behavior by approaching along the line \(x = \pi\text{.}\)
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Approaching \(0\) from the right (\(y \gt 0\)):
\begin{equation*} \lim_{y \to 0^+} \dfrac{\cos\lp \pi \rp}{\sin\lp y \rp} = \lim_{y \to 0^+} \dfrac{-1}{\sin\lp y \rp} = -\infty \end{equation*}
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Approaching \(0\) from the left (\(y \lt 0\)):
\begin{equation*} \lim_{y \to 0^-} \dfrac{\cos\lp \pi \rp}{\sin\lp y \rp} = \lim_{y \to 0^-} \dfrac{-1}{\sin\lp y \rp} = \infty \end{equation*}
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Since the function approaches different infinities depending on the direction of approach, the limit \(\text{does not exist}\text{.}\)
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14.2.41.
\(\ds \lim_{(x,y) \to (0,0)} \dfrac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1}\)
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Solution.
We rewrite the function by dividing and multiplying it by the conjugate of \(\sqrt{x^2 + y^2 + 1} - 1\) and use the identity \((a - b)(a + b) = a^2 - b^2\text{.}\) This gives
\begin{align*} \frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1} \amp= \frac{\lp x^2 + y^2 \rp \lp \sqrt{x^2 + y^2 + 1} + 1 \rp}{\lp\sqrt{x^2 + y^2 + 1} - 1)\rp\lp\sqrt{x^2 + y^2 + 1} + 1\rp} \\ \amp= \frac{\lp x^2 + y^2 \rp \lp \sqrt{x^2 + y^2 + 1} + 1 \rp}{x^2 + y^2 + 1 - 1} \\ \amp= \frac{\lp x^2 + y^2 \rp \lp \sqrt{x^2 + y^2 + 1} + 1 \rp}{x^2 + y^2} \\ \amp= \sqrt{x^2 + y^2 + 1} + 1 \end{align*}
This resulting function is continuous, hence we may compute the limit by substitution:
\begin{align*} \lim_{(x,y) \to (0,0)} \dfrac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1} \amp= \lim_{(x,y) \to (0,0)} \lp \sqrt{x^2 + y^2 + 1} + 1 \rp \\ \amp= \sqrt{0^2 + 0^2 + 1} + 1 \\ \amp= 2 \end{align*}
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