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Evaluate $\displaystyle{\lim_{x \rightarrow +\infty} \sec \left( \frac{1}{x} \right)}$ and provide a graphical interpretation.
$\displaystyle{\lim_{x \rightarrow +\infty} \sec \left( \frac{1}{x} \right) = \lim_{u \rightarrow 0} \sec u = \sec(0) = 1}$; Graphically, there is a horizontal asymptote of $y=1$ associated with the graph of $\sec(1/x)$.
Evaluate $\displaystyle{\lim_{x \rightarrow 0} e^{1/x}}$ and provide a graphical interpretation.
We must be careful here. As $x \rightarrow 0$, the behavior of $1/x$ depends upon the side from which we are approaching $0$. Thus, we first compute the two one-sided limits.
$$\lim_{x \rightarrow 0^-} \frac{1}{x} = -\infty \quad \textrm{ and } \quad \lim_{x \rightarrow 0^+} \frac{1}{x} = +\infty$$Now we can find the left and right limiting values corresponding to the original limit:
$$\lim_{x \rightarrow 0^-} e^{1/x} = \lim_{u \rightarrow -\infty} e^u = 0 \quad \textrm{ and } \lim_{x \rightarrow 0^+} e^{1/x} = \lim_{u \rightarrow \infty} e^u = +\infty$$Of course, as the left and right limits disagree,
$$\lim_{x \rightarrow 0} e^{1/x} \quad \textrm{does not exist}$$Graphically, the infinite limit on the right tells us there is a vertical asympotote at $x=0$. Although, in combination with the left limiting value of $0$ suggests the behavior might also be described as an "infinite gap discontinuity" at $x=0$.
Evaluate $\displaystyle{\lim_{x \rightarrow -\infty} \ln \left( \frac{1}{x^2} \right)}$ and provide a graphical interpretation.
Note that as $x \rightarrow -\infty$, the fraction $1/x^2$ gets very small (arbitrarily close to zero), but always remains positive due to the squaring. As such, we write $\displaystyle{\lim_{x \rightarrow -\infty} \frac{1}{x^2} = 0^+}$. From that point, things are direct:
$$\lim_{x \rightarrow -\infty} \ln \left( \frac{1}{x^2} \right) = \lim_{u \rightarrow 0^+} \ln u = -\infty$$Note, the limiting value is infinite. Had $x$ been approaching a finite value, this would normally suggest the presence of a vertical asymptote.
However, here $x$ is approaching $-\infty$ (i.e., we are looking at the behavior of the related function on the far left). Thus, in terms of a graphical interpretation, the function $y = \ln(1/x^2)$ is simply decreasing without bound as $x$ becomes sufficiently negative.
Suppose $\displaystyle{\delta(x) = \left\{ \begin{array}{ll} 1 & \textrm{ if } x = 0\\ 0 & \textrm{ otherwise } \end{array} \right.}$
Evaluate the following limits:
(a) $\displaystyle{\lim_{x \rightarrow 0} \delta(x)}$ (b) $\displaystyle{\lim_{x \rightarrow +\infty} \delta(1/x)}$ (c) $\displaystyle{\lim_{x \rightarrow 0} \delta(\cos(1/x))}$ (d) $\displaystyle{\lim_{x \rightarrow 0} \delta(\delta(x))}$
(a) $0$
(b) $\displaystyle{\lim_{x \rightarrow +\infty} \delta(1/x) = \lim_{u \rightarrow 0} \delta(u) = 0}$
(c) Here, as the behavior of $1/x$ as $x \rightarrow 0$ depends upon the side from which we are approaching zero, we split things into left and right limits:
$$\lim_{x \rightarrow 0^-} \frac{1}{x} = -\infty \quad \textrm{ and } \quad \lim_{x \rightarrow 0^+} \frac{1}{x} = +\infty$$Thus,
$$\lim_{x \rightarrow 0^-} \delta(\cos(1/x)) = \lim_{u \rightarrow -\infty} \delta(\cos u) \quad \textrm{ and } \quad \lim_{x \rightarrow 0^+} \delta(\cos(1/x)) = \lim_{u \rightarrow +\infty} \delta(\cos u)$$However, as $\cos u$ simply oscillates back and forth between $-1$ and $1$ as both $u \rightarrow -\infty$ and $u \rightarrow +\infty$, the input to $\delta$ is zero infinitely many times. For these inputs, $\delta = 1$.
For the rest (also infinitely many values), the input to $\delta$ is non-zero, making $\delta = 0$.
As such, there is no single value to which $\delta(\cos (1/x))$ gets arbitrarily near.
Consequently, $\displaystyle{\lim_{x \rightarrow 0} \delta(\cos(1/x))}$ does not exist.
(d) $\displaystyle{\lim_{x \rightarrow 0} \delta(\delta(x)) = \lim_{u \rightarrow 0} \delta(u) = 0}$
Evaluate $\displaystyle{\lim_{y \rightarrow \frac{1}{2}} \left[ \textrm{Arcsin } \sqrt{y} \right]^2}$ and provide a graphical interpretation.
$\displaystyle{\lim_{y \rightarrow \frac{1}{2}} \left[ \textrm{Arcsin } \sqrt{y} \right]^2 = \lim_{u \rightarrow \frac{\sqrt{2}}{2}} \left[ \textrm{Arcsin } u \right]^2 = \lim_{w \rightarrow \frac{\pi}{4}} w^2 = \frac{\pi^2}{16}}$
As this limiting value could have been evaluated by direct substitution of $y=1/2$, there is a point of continuity in the related graph at $(1/2,\pi^2/16)$.
Find the value of $\displaystyle{\lim_{x \rightarrow 1^+} e^{1/\ln x}}$ and state its graphical interpretation.
$\displaystyle{\lim_{x \rightarrow 1^+} e^{1/\ln x} = \lim_{u \rightarrow 0^+} e^{1/u} = \lim_{w \rightarrow +\infty} e^w = +\infty}$
Graphically, there is a vertical asymptote in the graph of $y = e^{1/\ln x}$ at $x = 1$.
Find the value of $\displaystyle{\lim_{x \rightarrow 1^-} e^{1/\ln x}}$ and state its graphical interpretation.
$\displaystyle{\lim_{x \rightarrow 1^-} e^{1/\ln x} = \lim_{u \rightarrow 0^-} e^{1/u} = \lim_{w \rightarrow -\infty} e^w = 0}$
Graphically, there is a hole (from the left) at $(1,0)$.
Find the value of $\displaystyle{\lim_{s \rightarrow 0} \textrm{Arctan } (1/s)}$ and state its graphical interpretation.
Note the behavior of $1/s$ as $s \rightarrow 0$ depends upon the side from which we approach $0$. As such, we consider the left and right limits separately:
$\displaystyle{\lim_{s \rightarrow 0^-} \textrm{Arctan } (1/s) = \lim_{u \rightarrow -\infty} \textrm{Arctan } u = -\pi/2 \quad \textrm{ and } \quad \lim_{s \rightarrow 0^+} \textrm{Arctan } (1/s) = \lim_{u \rightarrow +\infty} \textrm{Arctan } u = \pi/2}$As the left and right limits disagree, $\displaystyle{\lim_{s \rightarrow 0} \textrm{Arctan } (1/s)}$ does not exist.
Graphically, we have a gap discontinuity from $(0,-\pi/2)$ to $(0,\pi/2)$.
Find the value of $\displaystyle{\lim_{p \rightarrow \sqrt{\pi/3}} \csc^3 p^2}$ and state its graphical interpretation.
$\displaystyle{\lim_{p \rightarrow \sqrt{\pi/3}} \csc^3 p^2 = \lim_{u \rightarrow \pi/3} (\csc u)^3 = \lim_{w \rightarrow \frac{2\sqrt{3}}{3}} w^3 = \frac{8\sqrt{3}}{9}}$
Graphically, as this limit could be found by direct substitution we have a point of continuity in the related graph at $\displaystyle{\left( \sqrt{\frac{\pi}{3}},\frac{8\sqrt{3}}{9}\right)}$.
Find the value of $\displaystyle{\lim_{x \rightarrow -\infty} \cos(1/x)}$ and state its graphical interpretation.
$\displaystyle{\lim_{x \rightarrow -\infty} \cos(1/x) = \lim_{u \rightarrow 0} \cos u = \cos(0) = 1}$
Graphically, the graph of $y = \cos(1/x)$ has a horizontal asymptote on the left of $y=1$.
Evaluate $\displaystyle{\lim_{x \rightarrow +\infty} \frac{\cos x}{x}}$ and provide a graphical interpretation for this limit.
Since $x \rightarrow +\infty$, we may focus our attention only on $x \gt 0$.
Since $-1 \le \cos x \le 1$, we can conclude that $\displaystyle{\frac{-1}{x} \le \frac{\cos x}{x} \le \frac{1}{x}}$.
As $\displaystyle{\lim_{x \rightarrow +\infty} \frac{-1}{x} = 0}$ and $\displaystyle{\lim_{x \rightarrow +\infty} \frac{1}{x} = 0}$ (the same value), by the squeeze theorem we may conclude
$$\displaystyle{\lim_{x \rightarrow +\infty} \frac{\cos x}{x} = 0}$$Graphically, this means there is a horizontal asymptote of $y=0$ on the right side of the graph of $\displaystyle{y = \frac{\cos x}{x}}$.
Given $\displaystyle{f(x) = \left\{ \begin{array}{cc} -\sqrt{16-x^2}, & x \lt 0\\ 3, & x = 0\\ x^2 - 4, & 0 \lt x \le 3\\ 3+\sqrt{7-x}, & x \gt 3 \end{array} \right.}$
Find the following:
(a) $\displaystyle{\lim_{x \rightarrow -4} f(x)}$ (b) $\displaystyle{\lim_{x \rightarrow 0} f(x)}$ (c) $\displaystyle{\lim_{x \rightarrow 2} f(x)}$ (d) $\displaystyle{\lim_{x \rightarrow 3} f(x)}$ (e) $f(0)$
(a) $0$; (b) $-4$; (c) $0$; (d) $5$; (e) $3$
Use the squeeze theorem to show that $\displaystyle{\lim_{x \rightarrow +\infty} \frac{x^2-x}{1-x^2} = -1}$.
Hint: note that for $x \gt 1$, the following holds $\displaystyle{\frac{x^2-1}{1-x^2} \le \frac{x^2-x}{1-x^2} \le \frac{x^2-x}{-x^2}}$.