So far only the intuitive notion of a limit has been used, namely:
That notion can be put in terms of a formal definition as follows:
A visual way of thinking of this definition is shown in Figure [fig:limit] below:
Figure [fig:limit] says that for any interval around \(L\) on the \(y\)-axis, you will be able to find at least one small interval around \(x = a\) (but excluding \(a\)) on the \(x\)-axis that the function \(y=f(x)\) maps completely inside that interval on the \(y\)-axis. Choosing smaller intervals around \(L\) on the \(y\)-axis could force you to find smaller intervals around \(a\) on the \(x\)-axis.
In Figure [fig:limit], \(f(x)\) is made arbitrarily close to \(L\) (within any distance \(\epsilon > 0\)) by picking \(x\) sufficiently close to \(a\) (within some distance \(\delta > 0\)). Since \(0 < \abs{x - a} < \delta\) means that \(x = a\) itself is excluded, the solid dot at \((a,L)\) could even be a hollow dot. That is, \(f(a)\) does not have to equal \(L\), or even be defined; \(f(x)\) just needs to approach \(L\) as \(x\) approaches \(a\). Thus–perhaps counter-intuitively—the existence of the limit does not actually depend on what happens at \(x=a\) itself.
Example \(\PageIndex{1}\): limx
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Solution
Show that \(\displaystyle\lim_{x \to a} ~x = a\,\) for any real number \(a\).
Solution: Though the limit is obvious, the following “epsilon-delta” proof shows how to use the formal definition. The idea is to let \(\epsilon > 0\) be given, then “work backward” from the inequality \(\abs{f(x)-L} < \epsilon\) to get an inequality of the form \(\abs{x-a} < \delta\), where \(\delta > 0\) usually depends on \(\epsilon\). In this case the limit is \(L = a\) and the function is \(f(x) = x\), so since
\[\begin{aligned} \abs{f(x)-a} < \epsilon \quad&\Leftrightarrow\quad \abs{x-a} < \epsilon ~,\\ \intertext{then choosing $\delta = \epsilon$ means that} 0 < \abs{x-a} < \delta \quad&\Rightarrow\quad \abs{x-a} < \epsilon \quad\Rightarrow\quad \abs{f(x)-a} < \epsilon ~, \end{aligned} \nonumber \]
which by definition means that \(\displaystyle\lim_{x \to a} ~x = a\).
Calculating limits in this way might seem silly since—as in Example
Example \(\PageIndex{1}\): limx
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Solution
—it requires extra effort for a result that is obvious. The formal definition is used most often in proofs of general results and theorems. For example, the rules for limits—listed in Section 1.2—can be proved by using the formal definition.
Example \(\PageIndex{1}\): limsum
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Solution
Suppose that \(\displaystyle\lim_{x \to a} f(x)\) and \(\displaystyle\lim_{x \to a} g(x)\) both exist. Show that
\[\lim_{x \to a} ~(f(x) + g(x)) ~=~ \left(\lim_{x \to a} ~f(x)\right) ~+~ \left(\lim_{x \to a} ~g(x)\right) \nonumber \]
Solution: Let \(\displaystyle\lim_{x \to a} f(x) = L_1\) and \(\displaystyle\lim_{x \to a} g(x) = L_2\). The goal is to show that \(\displaystyle\lim_{x \to a} ~(f(x) + g(x)) = L_1 + L_2\). So let \(\epsilon > 0\). Then \(\epsilon/2 > 0\), and so by definition there exist numbers \(\delta_1 > 0\) and \(\delta_2 > 0\) such that
\[\begin{aligned} 0 < \abs{x - a} < \delta_1 \quad&\Rightarrow\quad \abs{f(x) ~-~ L_1} < \epsilon/2~\text{, and}\\ 0 < \abs{x - a} < \delta_2 \quad&\Rightarrow\quad \abs{g(x) ~-~ L_2} < \epsilon/2 ~.\end{aligned} \nonumber \]
Now let \(\delta = \min(\delta_1, \delta_2)\). Then \(\delta > 0\) and
\[\begin{aligned} 0 < \abs{x - a} < \delta \quad&\Rightarrow\quad 0 < \abs{x - a} < \delta_1 \quad\text{and}\quad 0 < \abs{x - a} < \delta_2\\ &\Rightarrow\quad \abs{f(x) ~-~ L_1} < \epsilon/2 \quad\text{and}\quad \abs{g(x) ~-~ L_2} < \epsilon/2\end{aligned} \nonumber \]
Since \(\abs{A+B} \le \abs{A} + \abs{B}\) for all real numbers \(A\) and \(B\), then
\[\abs{f(x) ~+~ g(x) ~-~ (L_1 + L_2)} ~=~ \abs{(f(x) ~-~ L_1) ~+~ (g(x) ~-~ L_2)} ~\le~ \abs{f(x) ~-~ L_1} ~+~ \abs{g(x) ~-~ L_2} \nonumber \]
and thus
\[0 < \abs{x-a} < \delta \quad\Rightarrow\quad \abs{f(x) ~+~ g(x) ~-~ (L_1 + L_2)} ~<~ \epsilon/2 ~+~ \epsilon/2 ~=~ \epsilon \nonumber \]
which by definition means that \(\displaystyle\lim_{x \to a} ~(f(x) + g(x)) = L_1 + L_2\).
The proofs of the other limit rules are similar.1 In general, using the formal definition will not be necessary for evaluating limits of specific functions—in many cases a simple analysis of the function is all that is needed, often from its graph.
Example \(\PageIndex{1}\): limsimple
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Solution
Evaluate \(\displaystyle\lim_{x \to 1}~f(x)\) for the following function:
\[f(x) ~=~ \begin{cases} ~x & \text{if } x > 1\\ ~2 & \text{if } x = 1\\ ~1 & \text{if } x < 1 \end{cases} \nonumber \]
Solution: From the graph of \(f(x)\) in Figure [fig:limsimple], it is clear that as \(x\) approaches \(1\) from the right (i.e. for \(x > 1\)) \(f(x)\) approaches \(1\) along the line \(y = x\), whereas as \(x\) approaches \(1\) from the left (i.e. for \(x < 1\)) \(f(x)\) approaches \(1\) along the horizontal line \(y = 1\). Thus, \(\displaystyle\lim_{x \to 1}~f(x) = 1~\).
Note that the limit did not depend on the value of \(f(x)\) at \(x = 1\).
As Example
Example \(\PageIndex{1}\): limsimple
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Solution
shows, what matters for a limit is what happens to the value of \(f(x)\) as \(x\) gets near \(a\), not at \(x=a\) itself. Figure [fig:lims](a) below shows how as \(x\) approaches \(a\), \(f(x)\) approaches a number different from \(f(a)\). Figure [fig:lims](b) shows that \(x=a\) does not even need to be in the domain of \(f(x)\), i.e. \(f(a)\) does not have to be defined. So it will not always be the case that \(\displaystyle\lim_{x \to a} ~f(x) = f(a)\).
In Example
Example \(\PageIndex{1}\): limsimple
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Solution
the direction in which \(x\) approached the number \(1\) did not affect the limit. But what if \(f(x)\) had approached different values depending on how \(x\) approached \(1\)? In that case the limit would not exist. The following definitions and notation for one-sided limits will make situations like that simpler to state.
The following statement follows immediately from the above definitions:
Example \(\PageIndex{1}\): onesidedlim
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Solution
Evaluate \(\displaystyle\lim_{x \to 0-}~f(x)\), \(\displaystyle\lim_{x \to 0+}~f(x)\), and \(\displaystyle\lim_{x \to 0}~f(x)\) for the following function:
\[f(x) ~=~ \begin{cases} ~x^2 & \text{if } x < 0\\ ~2 - x & \text{if } x \ge 0 \end{cases} \nonumber \]
Solution: From the graph of \(f(x)\) in Figure [fig:onesidedlim], it is clear that as \(x\) approaches \(0\) from the left (i.e. for \(x < 0\)) \(f(x)\) approaches \(0\) along the parabola \(y = x^2\), whereas as \(x\) approaches \(0\) from the right (i.e. for \(x > 0\)) \(f(x)\) approaches \(2\) along the line \(y = 2 - x\). Hence, \(\displaystyle\lim_{x \to 0-}~f(x) = 0~\) and \(\displaystyle\lim_{x \to 0+}~f(x) = 2~\). Thus, \(\displaystyle\lim_{x \to 0}~f(x) ~~\text{does not exist}~\) since the left and right limits do not agree at \(x=0\).
Example \(\PageIndex{1}\): limsin1overx
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Solution
Evaluate \(\displaystyle\lim_{x \to 0+}~\sin\,\left(\dfrac{1}{x}\right)~\).
Solution: For \(x > 0\) the function \(f(x) = \sin\,(1/x)\) is defined, and its graph is shown in Figure [fig:limsin1overx]. As \(x\) approaches \(0\) from the right, \(\sin\,(1/x)\) will be \(1\) for the numbers \(x = 2/\pi\), \(2/5\pi\), \(2/9\pi\), \(2/13\pi\), \(\ldots\) (which approach \(0\)), and \(\sin\,(1/x)\) will be \(-1\) for the numbers \(x = 2/3\pi\), \(2/7\pi\), \(2/11\pi\), \(2/15\pi\), \(\ldots\) (which also approach \(0\)). So as \(x\) approaches \(0\) from the right, \(\sin\,(1/x)\) will oscillate between \(1\) and \(-1\). Thus, \(\displaystyle\lim_{x \to 0+}~\sin\,\left(\dfrac{1}{x}\right) ~~\text{does not exist}~\).
So far only finite limits have been considered, that is, \(L = \displaystyle\lim_{x \to a}~f(x)\) where \(L\) is a real (i.e. finite) number. Define an infinite limit, with \(L = \infty\) or \(-\infty\), as follows:
The above definitions can be modified accordingly for one-sided limits. If \(\displaystyle\lim_{x \to a} ~f(x) = \infty\) or \(\displaystyle\lim_{x \to a} ~f(x) = -\infty\), then the line \(x=a\) is a vertical asymptote of \(f(x)\), and \(f(x)\) approaches the line \(x=a\,\) asymptotically. The formal definitions are rarely needed.
Example \(\PageIndex{1}\): lim1overx
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Solution
Evaluate \(\displaystyle\lim_{x \to 0}~\dfrac{1}{x}~\).
Solution: For \(x \ne 0\) the function \(f(x) = \frac{1}{x}\) is defined, and its graph
is shown in Figure [fig:lim1overx]. As \(x\) approaches \(0\) from the right, \(1/x\)
approaches \(\infty\), that is,
\[\lim_{x \to 0+}~\dfrac{1}{x} ~=~ \infty ~. \nonumber \]
As \(x\) approaches \(0\) from the left, \(1/x\) approaches \(-\infty\), that is,
\[\lim_{x \to 0-}~\dfrac{1}{x} ~=~ -\infty ~. \nonumber \]
Since the right limit and the left limit are not equal, then \(\displaystyle\lim_{x \to 0}~\dfrac{1}{x} ~\text{does not exist}\).
Note that the \(y\)-axis (i.e. the line \(x = 0\)) is a vertical asymptote for \(f(x) = \frac{1}{x}\).
Example \(\PageIndex{1}\): lim1overx2
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Solution
Evaluate \(\displaystyle\lim_{x \to 0}~\dfrac{1}{x^2}~\).
Solution: For \(x \ne 0\) the function \(f(x) = \frac{1}{x^2}\) is defined, and its graph
is shown in Figure [fig:lim1overx2]. As \(x\) approaches \(0\) from either the right
or the left, \(1/x^2\) approaches \(\infty\), that is,
\[\lim_{x \to 0+}~\dfrac{1}{x^2} ~=~ \infty ~=~ \lim_{x \to 0-}~\dfrac{1}{x^2} ~. \nonumber \]
Since the right limit and the left limit both equal \(\infty\), then \(\displaystyle\lim_{x \to 0}~\dfrac{1}{x^2} ~=~ \infty\).
Note that the \(y\)-axis (i.e. the line \(x = 0\)) is a vertical asymptote for \(f(x) = \frac{1}{x^2}\).
In the limit \(\displaystyle\lim_{x \to a}~f(x)\) so far only real values of \(a\) have been considered. However, \(a\) could be either \(\infty\) or \(-\infty\):
The above definitions can be modified accordingly for \(L\) replaced by either \(\infty\) or \(-\infty\).
One way to interpret the statement \(\displaystyle\lim_{x \to \infty} ~f(x) ~=~ L\) is: the long-term behavior of \(f(x)\) is to approach a steady-state at \(L\). If \(\displaystyle\lim_{x \to \infty} ~f(x) = L\) or \(\displaystyle\lim_{x \to -\infty} ~f(x) = L\), then the line \(y=L\) is a horizontal asymptote of \(f(x)\), and \(f(x)\) approaches the line \(y=L\,\) asymptotically. Again, for most limits of specific functions, only the intuitive notions are needed.
Example \(\PageIndex{1}\): lim1overxandx2inf
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Solution
From Figures [fig:lim1overx] and [fig:lim1overx2], it is clear that
\[\lim_{x \to \infty}~\dfrac{1}{x} ~=~ 0 ~=~ \lim_{x \to -\infty}~\dfrac{1}{x} \qquad\text{and}\qquad \lim_{x \to \infty}~\dfrac{1}{x^2} ~=~ 0 ~=~ \lim_{x \to -\infty}~\dfrac{1}{x^2} \nonumber \]
Note that the \(x\)-axis (i.e. the line \(y = 0\)) is a horizontal asymptote for \(f(x) = \frac{1}{x}\) and \(f(x) = \frac{1}{x^2}\).
Some limits are obvious—you can use them for calculating other limits:
A related notion is that of Big O notation (that is the capital letter O, not a zero):
For example, obviously \(2x^3 = O(x^3)\), by picking \(M = 2\), with \(x_0\) any positive number. In general, \(f(x) = O(g(x))\) means that \(f\) exhibits the same long-term behavior as \(g\), up to a constant multiple. You can think of \(g\) as the more basic “type” of function that describes \(f\), as far as long-term behavior.
Example \(\PageIndex{1}\): bigOexmp
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Solution
Show that \(5x^4 - 2 = O(x^4)\).
Solution: First, recall from algebra that \(\abs{a + b} \le \abs{a} + \abs{b}\) for all real numbers \(a\) and \(b\). Thus,
\[\Abs{5x^4 - 2} ~\le~ \Abs{5x^4} ~+~ \abs{-2} ~=~ 5\,\Abs{x^4} ~+~ 2 \nonumber \]
for all \(x\). So since \(\Abs{x^4} = x^4 \ge 1\) for all \(x \ge 1\), then
\[\Abs{5x^4 - 2} ~\le~ 5\,\Abs{x^4} ~+~ 2 ~\le~ 5\,\Abs{x^4} ~+~ 2\,\Abs{x^4} \quad\Rightarrow\quad \Abs{5x^4 - 2} ~\le~ 7\,\Abs{x^4} ~~\text{for all $x \ge 1$,} \nonumber \]
which shows that \(5x^4 - 2 = O(x^4)\), with \(M= 7\) and \(x_0 = 1\).
Some limits need algebraic manipulation before they can be evaluated.
Example \(\PageIndex{1}\): liminfminusinf
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Solution
Evaluate \(\displaystyle\lim_{x \to \infty}~\left(\sqrt{x + 1} \,-\, \sqrt{x}\right)~\).
Solution: Note that both \(\sqrt{x + 1}\) and \(\sqrt{x}\) approach \(\infty\) as \(x\) goes to \(\infty\), resulting in a limit of the form \(\infty - \infty\). This is an example of an indeterminate form, which can equal anything (as will be discussed shortly); it does not have to equal \(0\) (i.e. the \(\infty\)’s do not necessarily “cancel out”). The trick here is to use the conjugate of \(\sqrt{x + 1} \,-\, \sqrt{x}\), so that
\[\lim_{x \to \infty}~\left(\sqrt{x + 1} \,-\, \sqrt{x}\right) ~=~ \lim_{x \to \infty}~\left(\sqrt{x + 1} \,-\, \sqrt{x}\right) \cdot \frac{\sqrt{x + 1} \,+\, \sqrt{x}}{\sqrt{x + 1} \,+\, \sqrt{x}} ~=~ \lim_{x \to \infty}~\frac{(x + 1) ~-~ x}{\sqrt{x + 1} \,+\, \sqrt{x}} ~=~ \lim_{x \to \infty}~\frac{1}{\sqrt{x + 1} \,+\, \sqrt{x}} ~=~ 0 \nonumber \]
since the numerator is \(1\) and both terms in the sum in the denominator approach \(\infty\) (i.e. \(\frac{1}{\infty} = 0\)).
Some other indeterminate forms are \(\infty/\infty\), \(0/0\) and \(\infty \cdot 0\). How would you handle such limits? One way is to use L’Hôpital’s Rule2; a simplified form is stated below:
Example \(\PageIndex{1}\): limxexpx
Add text here.
Solution
Evaluate \(\displaystyle\lim_{x \to \infty}~\dfrac{2x - 1}{e^x}~\).
Solution: This limit is of the form \(\infty/\infty\):
\[\begin{aligned} \lim_{x \to \infty}~\dfrac{2x - 1}{e^x} ~&\to~ \frac{\infty}{\infty}\\ &=~ \lim_{x \to \infty}~\dfrac{2}{e^x} \quad\text{by L'H\^{o}pital's Rule}\\ &=~ 0\end{aligned} \nonumber \]
since the numerator is \(2\) and \(e^x \to \infty\) as \(x \to \infty\).
Note that one way of interpreting the limit being \(0\) is that \(e^x\) grows much faster than \(2x - 1\). In fact, using L’Hôpital’s Rule it can be shown that \(e^x\) grows much faster than any polynomial, i.e. exponential growth outstrips polynomial growth.
Example \(\PageIndex{1}\): limxlnx
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Solution
Evaluate \(\displaystyle\lim_{x \to \infty}~\dfrac{x}{\ln\,x}~\).
Solution: This limit is of the form \(\infty/\infty\):
\[\begin{aligned} \lim_{x \to \infty}~\dfrac{x}{\ln\,x} ~&\to~ \frac{\infty}{\infty}\\ &=~ \lim_{x \to \infty}~\dfrac{1}{\frac{1}{x}} \quad\text{by L'H\^{o}pital's Rule}\
\[6pt] &=~ \lim_{x \to \infty}~x ~=~ \infty\end{aligned} \nonumber \]
Note that one way of interpreting the limit being \(\infty\) is that \(x\) grows much faster than \(\ln\,x\). In fact, using L’Hôpital’s Rule it can be shown that any polynomial grows much faster than \(\ln\,x\), i.e. polynomial growth outstrips logarithmic growth.
Example \(\PageIndex{1}\): limzeroinf
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Solution
Evaluate \(\displaystyle\lim_{x \to \infty}~x e^{-2x}~\).
Solution: This limit is of the form \(\infty \cdot 0\), which can be converted to \(\infty/\infty\):
\[\begin{aligned} \lim_{x \to \infty}~x e^{-2x} ~=~ \lim_{x \to \infty}~\frac{x}{e^{2x}} ~&\to~ \frac{\infty}{\infty}\\ &=~ \lim_{x \to \infty}~\dfrac{1}{2e^{2x}} \quad\text{by L'H\^{o}pital's Rule}\\ &=~ 0\end{aligned} \nonumber \]
Note that the limit is another consequence of exponential growth outstripping polynomial growth.
Example \(\PageIndex{1}\): limratpoly
Add text here.
Solution
Evaluate \(\displaystyle\lim_{x \to \infty}~\dfrac{2x^2 ~-~ 7x ~-~ 5}{3x^2 ~+~ 2x ~-~ 1}~\).
Solution: This limit is of the form \(\infty/\infty\):
\[\begin{aligned} \lim_{x \to \infty}~\dfrac{2x^2 ~-~ 7x ~-~ 5}{3x^2 ~+~ 2x ~-~ 1} ~&\to~ \frac{\infty}{\infty}\
\[4pt] &=~ \lim_{x \to \infty}~\dfrac{4x ~-~ 7}{6x ~+~ 2} \quad\text{by L'H\^{o}pital's Rule}\
\[6pt] ~&\to~ \frac{\infty}{\infty} \quad\text{, so use L'H\^{o}pital's Rule again}\
\[6pt] &=~ \frac{4}{6} ~=~ \frac{2}{3}\end{aligned} \nonumber \]
Note that the limit ended up being the ratio of the leading coefficients of the polynomials in the numerator and denominator of the original limit. Note also that the lower-order terms (degree less than 2) ended up not mattering. In general you can always discard the lower-order terms when taking the limit of a ratio of polynomials (i.e. a rational function).
Example \(\PageIndex{1}\): lim1minuscos
Add text here.
Solution
Evaluate \(\displaystyle\lim_{x \to 0}~\dfrac{1 ~-~ \cos\,x}{x}~\).
Solution: This limit is of the form \(0/0\):
\[\begin{aligned} \lim_{x \to 0}~\dfrac{1 ~-~ \cos\,x}{x} ~&\to~ \frac{0}{0}\
\[4pt] &=~ \lim_{x \to 0}~\dfrac{\sin\,x}{1} \quad\text{by L'H\^{o}pital's Rule}\
\[6pt] &=~ \frac{\sin 0}{1} ~=~ 0\end{aligned} \nonumber \]
There is an intuitive justification for L’Hôpital’s Rule: since the limit \(\lim_{x \to a}\;\frac{f(x)}{g(x)}\) uses a ratio to compare how \(f\) changes relative to \(g\) as \(x\) approaches \(a\), then it is really the rates of change of \(f\) and \(g\)—namely \(f'\) and \(g'\), respectively—that are being compared in that ratio, which is what L’Hôpital’s Rule says.
The following result provides another way to calculate certain limits:
Intuitively, the Squeeze Theorem says that if one function is “squeezed” between two functions approaching the same limit, then the function in the middle must also approach that limit. The theorem also applies to one-sided limits (\(x \to a+\;\) or \(\;x \to a-\)).
Example \(\PageIndex{1}\): squeeze
Add text here.
Solution
Evaluate \(\displaystyle\lim_{x \to \infty}~\dfrac{\sin\,x}{x}~\).
Solution: Since \(-1 \;\le\; \sin\,x \;\le\; 1\) for all \(x\), then dividing all parts of those inequalities by \(x > 0\) yields
\[-\frac{1}{x} ~\le~ \frac{\sin\,x}{x} ~\le~ \frac{1}{x} ~~\text{for all $x > 0$} \quad\Rightarrow\quad \lim_{x \to \infty}~\frac{\sin\,x}{x} ~=~ 0 \nonumber \]
by the Squeeze Theorem, since \(\displaystyle\lim_{x \to \infty}~\frac{-1}{x} ~=~ 0 ~=~ \displaystyle\lim_{x \to \infty}~\frac{1}{x}\).
[sec3dot2]
For Exercises 1-18 evaluate the given limit.
4
\(\displaystyle\lim_{x \to 2}~ \dfrac{x^2 + 3x - 10}{x^2 - x - 2}\)
\(\displaystyle\lim_{x \to \infty}~ \dfrac{x^2 + 3x - 10}{2x^2 - x - 2}\)
\(\displaystyle\lim_{x \to \infty}~ \dfrac{x^2 + 3x - 10}{2x^3 - x - 2}\)
\(\displaystyle\lim_{x \to \infty}~ \dfrac{x^3 + 3x - 10}{2x^2 - x - 2}\)
4
\(\displaystyle\lim_{x \to \pi/2}~ \dfrac{\cos\,x}{x - \pi/2}\vphantom{\dfrac{x^2}{e^x}}\)
\(\displaystyle\lim_{x \to \infty}~ \dfrac{x^2}{e^x}\)
\(\displaystyle\lim_{x \to -\infty}~ x^2 e^x\vphantom{\dfrac{x^2}{e^x}}\)
\(\displaystyle\lim_{x \to 0+}~ \dfrac{\ln\,x}{e^{1/x}}\vphantom{\dfrac{x^2}{e^x}}\)
4
\(\displaystyle\lim_{x \to 0}~ \dfrac{\tan\,x ~-~ x}{x ~-~ \sin\,x}\vphantom{\dfrac{\sin\,3x}{\sin\,4x}}\)
\(\displaystyle\lim_{x \to 0}~ \dfrac{\sin\,3x}{\sin\,4x}\)
\(\displaystyle\lim_{x \to \infty}~ \dfrac{\cos\,x}{x}\vphantom{\dfrac{\sin\,3x}{\sin\,4x}}\)
\(\displaystyle\lim_{x \to 0}~ x\,\sin\,\left(\dfrac{1}{x}\right)\vphantom{\dfrac{\sin\,3x}{\sin\,4x}}\)
3
\(\displaystyle\lim_{x \to 0}~ \dfrac{\ln\,(1-x) ~-~ \sin^2 x}{1 ~-~ \cos^2 x}\)
\(\displaystyle\lim_{x \to 1}~ \left(\dfrac{1}{\ln\,x} ~-~ \dfrac{1}{x-1}\right)\vphantom{\dfrac{\ln\,(1-x) ~-~ \sin^2 x}{1 ~-~ \cos^2 x}}\)
\(\displaystyle\lim_{x \to \pi/2}~ (\sec\,x ~-~ \tan\,x)\vphantom{\dfrac{\ln\,(1-x) ~-~ \sin^2 x}{1 ~-~ \cos^2 x}}\)
3
\(\displaystyle\lim_{x \to 0+}~ \dfrac{e^{-1/x}}{x}\)
\(\displaystyle\lim_{x \to \infty}~ \left(\sqrt{x^2 + 4} ~-~ x\right)\vphantom{\dfrac{e^{-1/x}}{x}}\)
\(\displaystyle\lim_{x \to 0}~ \dfrac{\cot\,x}{\csc\,x}\vphantom{\dfrac{e^{-1/x}}{x}}\)
[[1.]]
The famous “twin paradox,” a result of Einstein’s special theory of relativity, says that if one of a pair of twins leaves the earth in a rocket traveling at a high speed, then he will be younger than his twin upon returning to earth.3 This is due to the phenomenon of time dilation, which says that a clock moving with a speed \(v\) relative to a clock at rest in some inertial reference frame counts time slower relative to the clock at rest, by a factor of
\[\gamma = \dfrac{1}{\sqrt{1 - \beta^2}} ~, \nonumber \]
called the Lorentz factor, where \(\beta = \frac{v}{c}\) is the fraction of the speed of light \(c\) at which the clock is moving (\(c \approx 2.998 \times 10^8\) m/sec). Notice that \(0 \le \beta < 1\) (why?). For example, a clock moving at half the speed of light, so that \(\beta = 0.5\), would have \(\gamma = 1.1547\), meaning that the clock runs about \(15.47\%\) slower than the clock on earth.
- Evaluate \(\displaystyle\lim_{\beta \to 1-}~\gamma~\). What is the physical interpretation of this limit?
- Suppose an astronaut and his twin just turned 30 years old when the astronaut leaves earth on a high-speed journey through space. Upon returning to earth the astronaut is 35 and his twin is 70. At roughly what fraction of the speed of light must the astronaut have been traveling?
Show that \(\,\displaystyle\lim_{x \to \infty} ~\dfrac{p(x)}{e^x} = 0~\) for all polynomials \(p(x)\) of degree \(n \ge 1\) with a positive leading coefficient.
Show that \(\,\displaystyle\lim_{x \to \infty} ~\dfrac{p(x)}{\ln\,x} = \infty~\) for all polynomials \(p(x)\) of degree \(n \ge 1\) with a positive leading coefficient.
2
Show that \(5x^3 + 6x^2 - 4x + 3 = O(x^3)\vphantom{\dfrac{2x^2 + 1}{x + 1}}\).
Show that \(\dfrac{2x^2 + 1}{x + 1} = O(x)\). (Hint: Consider \(x \ge 1\))
Call \(h(x)\) an infinitesimal function as \(x \to a\,\) if \(\,\displaystyle\lim_{x \to a} ~h(x) = 0\). That is, an infinitesimal function approaches zero near some point. Prove the following result, where \(a\) and \(L\) are real numbers:
\[\lim_{x \to a}~f(x) ~=~ L \quad\Leftrightarrow\quad f(x) = L + h(x) \text{ for all $x$, where $h(x)$ is an infinitesimal function as $x \to a$} \nonumber \]
