Andrei Nikolaevich Schilder (1861-1919) was a Russian landscape painter.
The limit of a real-valued function \(f:\mathbb{R}\rightarrow\mathbb{R}\) is denoted by
\[\lim_{x\rightarrow p}f(x)=a,\]
which means that, given any \(\epsilon >0\), there exists \(\delta >0\) such that \(0<|x-p|<\delta\) implies \(|f(x)-a|<\epsilon\). We also recall that \(f\) is continuous at \(p\) when
\[\lim_{x\rightarrow p}f(x)=f(p).\]
In general, we are going to consider the limits of functions in metric space. The concepts of metric space, open ball and accumulation point can refer to the page point set topology in metric space. Now, we consider two metric spaces \((S,d_{S})\) and \((T,d_{T})\). Let \(A\) be a subset of \(S\) and consider the function \(f:(A,d_{S})\rightarrow (T,d_{T})\).
Definition. Let \(p\) be an accumulation point of \(A\), and let \(b\in T\). We write
\begin{equation}{\label{maeq433}}\tag{1}\lim_{x\rightarrow p}f(x)=b\end{equation}
when, given any \(\epsilon >0\), there exists \(\delta>0\) such that \(d_{S}(x,p)<\delta\) for \(x\in A\) implies \(d_{T}(f(x),b)<\epsilon\). \(\sharp\)
Remark. The definition of (\ref{maeq433}) can be formulated in terms of open balls. We see that \(d_{S}(x,p)<\delta\) for \(x\in A\) means \(x\in B_{S}(p;\delta )\cap A\neq\emptyset\), and that \(d_{T}(f(x),b)<\epsilon\) means \(f(x)\in B_{T}(b;\epsilon )\). The limit of function in (\ref{maeq433}) holds true if and only if, given any open ball \(B_{T}(b;\epsilon )\), there exists an open ball \(B_{S}(p;\delta )\) such that \(x\in B_{S}(p;\delta )\cap A\neq\emptyset\) implies \(f(x)\in B_{T}(b;\epsilon )\).
\begin{equation}{\label{map434}}\tag{2}\mbox{}\end{equation}
Proposition \ref{map434}. Let \(p\) be an accumulation point of \(A\), and let \(b\in T\). Then, the limit of function in (\ref{maeq433}) holds true if and only if
\[\lim_{n\rightarrow\infty}f(x_{n})=b\]
for every sequence \(\{x_{n}\}_{n=1}^{\infty}\) of points in \(A\setminus{p}\) satisfying
\[\lim_{n\rightarrow\infty}x_{n}=p.\]
Proof. Suppose that the limit of function in (\ref{maeq433}) holds true. Given any \(\epsilon >0\), there exists \(\delta>0\) such that \(x\in A\) and \(d_{S}(x,p)<\delta\) implies \(d_{T}(f(x),b)<\epsilon\). Since \(p\) is an accumulation point of \(A\), we can take a sequence \(\{x_{n}\}_{n=1}^{\infty}\) of points in \(A\setminus{p}\) satisfying
\[\lim_{n\rightarrow\infty}x_{n}=p.\]
Then, there exists an integer \(N\) such that \(n\geq N\) implies \(d_{S}(x_{n},p)<\delta\). Therefore, we obtain \(d_{T}(f(x_{n}),b)<\epsilon\) for \(n\geq N\). This says that the sequence \(\{f(x_{n})\}_{n=1}^{\infty}\) converges to \(b\).
For the converse, suppose that every sequence \(\{x_{n}\}_{n=1}^{\infty}\) of points in \(A\setminus{p}\) converges to \(p\), and that the limit of function in (\ref{maeq433}) does not hold true. We are going to lead to a contradiction. By definition, for some \(\epsilon >0\) and every \(\delta >0\), there exists a point \(x\in A\) (which depends on \(\delta\)) satisfying \(0<d_{S}(x,p)<\delta\) but \(d_{T}(f(x),b)\geq\epsilon\). We take \(\delta =1/n\). Then, we obtain a sequence of points \(\{x_{n}\}_{n=1}^{\infty}\) in \(A\setminus{p}\) satisfying \(0<d_{S}(x_{n},p)<1/n\) but \(d_{T}(f(x_{n}),b)\geq\epsilon\), which contradicts the convergence \(f(x_{n})\rightarrow b\) as \(x_{n}\rightarrow p\). This completes the proof. \(\blacksquare\)
Proposition. The limit of a function is unique.
Proof. The desired result can be realized by using Proposition \ref{map434} and the uniqueness of limits of sequences.
Proposition. Given a real-valued function \(f:S\rightarrow\in\mathbb{R}\) defined on a subset \(S\) of \(\mathbb{R}\), suppose that the limit \(\lim_{x\rightarrow p}f(x)\) exists. Then, we have
\[\lim_{x\rightarrow p}|f(x)|=\left |\lim_{x\rightarrow p}f(x)\right |.\]
Proof. Let
\[L=\lim_{x\rightarrow p}f(x).\]
Given any \(\epsilon >0\), there exists \(\delta>0\) satisfying \(|x-p|<\delta\) implies \(|f(x)-L|<\epsilon\). Since
\[\left ||f(x)|-|L|\right |\leq |f(x)-L|,\]
it follows that \(|x-p|<\delta\) implies
\[\left ||f(x)|-|L|\right |\leq |f(x)-L|<\epsilon,\]
which says
\[\lim_{x\rightarrow p}|f(x)|=|L|.\]
This completes the proof. \(\blacksquare\)
Example. We will show that the following function
\[f(x,y)=\frac{xy+y^{3}}{x^{2}+y^{2}}\]
does not have a limit at \((0,0)\). Note that \(f\) is not defined at \((0,0)\). Along the \(x\)-axis, i.e., \(y=0\), we have \(f(x,y)=f(x,0)=0\) and
\[\lim_{x\rightarrow 0} f(x,0)=0.\]
Along the \(y\)-axis, i.e., \(x=0\), we have \(f(x,y)=f(0,y)=y\) and
\[\lim_{y\rightarrow 0}f(0,y)=0.\]
Along the line \(y=2x\), we have
\begin{align*} f(x,y) & =f(x,2x)\\ & =\frac{2x^{2}+8x^{3}}{x^{2}+4x^{2}}\\ & =\frac{2}{5}+\frac{8}{5}x\rightarrow\frac{2}{5}\mbox{ as }x\rightarrow 0.\end{align*}
Along the line \(y=-x\), we also can verify
\[f(x,y)\rightarrow -\frac{1}{2}\mbox{ as }(x,y)\rightarrow (0,0).\]
Therefore, not all paths to \((0,0)\) yield the same limiting value. It follows that \(f\) does not have a limit at \((0,0)\). \(\sharp\)
Example. Show that the function
\[g(x,y)=\frac{x^{2}y}{x^{4}+y^{2}}\]
has limiting value \(0\) as \((x,y)\rightarrow (0,0)\) along any line through the origin. However, the limit \(\lim_{(x,y)\rightarrow (0,0)} g(x,y)\) still does not exist. Along the coordinate axes, we can verify
\[g(x,y)\rightarrow 0\mbox{ as }(x,y)\rightarrow (0,0).\]
Let \(y=mx\). Then, we have
\begin{align*} g(x,y) & =g(x,mx)\\ & =\frac{mx^{3}}{x^{4}+m^{2}x^{2}}\\ & =\frac{mx}{x^{2}+m^{2}}\end{align*}
and
\[\lim_{x\rightarrow 0} g(x,mx)=\lim_{x\rightarrow 0}\frac{mx}{x^{2}+m^{2}}=0.\]
Therefore, along any line through the origin, we have
\[g(x,y)\rightarrow 0\mbox{ as }(x,y)\rightarrow 0.\]
Suppose that \((x,y)\rightarrow (0,0)\) along the parabola \(y=x^{2}\). Then, we have
\begin{align*} g(x,y) & =g(x,x^{2})\\ & =\frac{x^{4}}{x^{4}+x^{4}}\\ & =\frac{1}{2}\end{align*}
and
\[\lim_{x\rightarrow 0} g(x,x^{2})=\frac{1}{2}.\]
Since not all paths to \((0,0)\) yield the same limiting value, we conclude that \(g\) does not have a limit at \((0,0)\). \(\sharp\)
Example. Given a two-valued function defined on \(\mathbb{R}\) by
\[f(x)=\left\{\begin{array}{ll} 0 & \mbox{if \(x\in\mathbb{Q}\)}\\ 1 & \mbox{if \(x\not\in\mathbb{Q}\)},\end{array}\right .\]
we are going to claim that the limit \(\lim_{x\rightarrow p}f(x)\) does not exists for any \(p\in\mathbb{R}\). Assume that the limit \(\lim_{x\rightarrow p}f(x)=L\) exists. Given \(\epsilon=1/3\), there exists \(\delta>0\) satisfying \(|x-p|<\delta\) implies
\[|f(x)-L|<\epsilon=\frac{1}{3}.\]
Using the denseness of \(\mathbb{Q}\) in \(\mathbb{R}\), we consider two cases as follows.
- Suppose that \(x\in\mathbb{Q}\) satisfying \(|x-p|<\delta\). Then, we have \[\epsilon >|f(x)-L|=|0-L|.\]
- Suppose that \(x\not\in\mathbb{Q}\) satisfying \(|x-p|<\delta\). Then, we have \[\epsilon >|f(x)-L|=|1-L|.\]
Now, we have
\begin{align*} 1 & =|0-1|\\ & \leq |0-L|+|L-1|\\ & <\epsilon+\epsilon\\ & =2\epsilon\\ & =\frac{2}{3}.\end{align*}
This contradiction says that the limit \(\lim_{x\rightarrow p}f(x)\) does not exists for any \(p\in\mathbb{R}\). \(\sharp\)
Example. Given a real-valued function defined on \(\mathbb{R}\) by
\[f(x)=\left\{\begin{array}{ll} x-1 & \mbox{if \(x\in\mathbb{Q}\)}\\ 5-x & \mbox{if \(x\not\in\mathbb{Q}\)},\end{array}\right .\]
we want to study its limits. We first have
\[\lim_{x\rightarrow 3}(x-1)=2=\lim_{x\rightarrow 3}(5-x).\]
Given any \(\epsilon >0\), there exist \(\delta_{1}>0\) and \(\delta_{2}>0\) such that \(|x-3|<\delta_{1}\) implies \(|(x-1)-2|<\epsilon\), and that \(|x-3|<\delta_{2}\) implies \(|(5-x)-2|<\epsilon\). Let \(\delta=\min{\delta_{1},\delta_{2}}\). Then, for \(|x-3|<\delta\), we have \(|(x-1)-2|<\epsilon\) and \(|(5-x)-2|<\epsilon\). According to the denseness of \(\mathbb{Q}\) in \(\mathbb{R}\), we consider the following two cases.
- Suppose that \(x\in\mathbb{Q}\) satisfying \(|x-3|<\delta\). We have \[|f(x)-2|=|(x-1)-2|<\epsilon.\]
- Suppose that \(x\not\in\mathbb{Q}\) satisfying \(|x-3|<\delta\). We have \[|f(x)-2|=|(5-x)-2|<\epsilon.\]
Therefore, we conclude that \(|x-3|<\delta\) implies \(|f(x)-2|<\epsilon\), which says
\[\lim_{x\rightarrow 3}f(x)=2.\]
Next, we want to claim that the limit \(\lim_{x\rightarrow a}f(x)\) does not exist for \(a\neq 3\). Assumes that the limit
\[L=\lim_{x\rightarrow a}f(x)\]
exists. Given \(\epsilon=|3-a|/3\), there exists \(\delta>0\) such that \(|x-a|<\delta\) implies
\[|f(x)-L|<\epsilon=\frac{|3-a|}{3}.\]
Let \(\delta^{*}=\min{\delta,\epsilon}\). We consider the following two cases.
- Suppose that \(x\in\mathbb{Q}\) satisfying \(|x-a|<\delta^{*}\). Then, we have \begin{align*} |f(x)-L| & =|(x-1)-L|\\ & <\epsilon,\end{align*} which implies \begin{align*} & |a-1-L|\\ & \quad\leq |a-x|+|x-1-L|\\ & \quad <\delta^{*}+\epsilon\\ & \quad\leq 2\epsilon.\end{align*}
- Suppose that \(x\not\in\mathbb{Q}\) satisfying \(|x-a|<\delta^{*}\). Then, we have \begin{align*} |f(x)-L| & =|(5-x)-L|\\ & <\epsilon,\end{align*} which implies \begin{align*} & |5-a-L|\\ & \quad\leq |5-x-L+|x-a|\\ & \quad<\epsilon+\delta^{*}\\ & \quad\leq 2\epsilon.\end{align*}
Therefore, we obtain
\begin{align*} 2|3-a| & =|6-2a|\\ & =|(5-a-L)-(a-1-L)|\\ & \leq |5-a-L|+|a-1-L|\\ & \leq 2\epsilon+2\epsilon\\ & =4\epsilon\\ & =\frac{4|3-a|}{3},\end{align*}
which implies
\[|3-a|\leq\frac{2|3-a|}{3}.\]
This contradiction says that the limit \(\lim_{x\rightarrow a}f(x)\) does not exist for \(a\neq 3\). \(\sharp\)
Theorem. (Squeeze Theorem).
Let \(f,g,h\) be three real-valued functions defined on a subset \(S\) of \(\mathbb{R}\) satisfying
\[f(x)\leq g(x)\leq h(x)\]
for all \(x\in S\). Suppose that
\[\lim_{x\rightarrow p}f(x)=\lim_{x\rightarrow p}h(x)=L.\]
Then, we have
\[\lim_{x\rightarrow p}g(x)=L.\]
Proof. Given any \(\epsilon>0\), there exist \(\delta_{1}>0\) and \(\delta_{2}>0\) such that \(|x-p|<\delta_{1}\) implies \(|f(x)-L|<\epsilon\), and that \(|x-p|<\delta_{2}\) implies \(|h(x)-L|<\epsilon\). Let \(\delta=\min{\delta_{1},\delta_{2}}\). For \(x\in S\) satisfying \(|x-p|<\delta\), we have
\[L-\epsilon<f(x)<L+\epsilon\]
and
\[L-\epsilon<h(x)<L+\epsilon,\]
which implies
\begin{align*} L-\epsilon & <f(x)\leq g(x) & \leq h(x) & <L+\epsilon.\end{align*}
Therefore, we obtain \(|g(x)-L|<\epsilon\). This completes the proof. \(\blacksquare\)
Theorem. Let \(f\) and \(g\) be two real-valued functions defined on \(\mathbb{R}^{n}\). Suppose that
\[\lim_{{\bf x}\rightarrow {\bf x}_{0}} f({\bf x})=L\]
and
\[\lim_{{\bf x}\rightarrow {\bf x}_{0}} g({\bf x})=M.\]
Then, we have
\[\lim_{{\bf x}\rightarrow {\bf x}_{0}}[f({\bf x})+g({\bf x})]=L+M,\]
\[\lim_{{\bf x}\rightarrow {\bf x}_{0}}[f({\bf x})g({\bf x})]=LM,\]
\[\lim_{{\bf x}\rightarrow {\bf x}_{0}}[\alpha f({\bf x})]=\alpha L,\]
where \(\alpha\) is a real number. We also have
\[\lim_{{\bf x}\rightarrow {\bf x}_{0}}[f({\bf x})/g({\bf x})]=L/M\]
provided \(M\neq 0\).
Proof. The proofs are left as exercises.
Example. Let \(f\) and \(g\) be two real-valued functions defined on a subset \(S\) of \(\mathbb{R}\) such that the limits
\[L_{f}=\lim_{x\rightarrow p}f(x)\]
and
\[L_{g}=\lim_{x\rightarrow p}g(x)\]
exist. Then, we have
\begin{align*} & \lim_{x\rightarrow p}\max\left\{f(x),g(x)\right\}\\ & \quad =\lim_{x\rightarrow p} \frac{f(x)+g(x)+|f(x)-g(x)|}{2}\\ & \quad =\frac{L_{f}+L_{g}+|L_{f}-L_{g}|}{2}\\ & \quad =\max\left\{L_{f},L_{g}\right\}\\ & \quad =\max\left\{\lim_{x\rightarrow p}f(x),\lim_{x\rightarrow p}g(x)\right\}. \end{align*}
and
\begin{align*} & \lim_{x\rightarrow p}\min\left\{f(x),g(x)\right\}\\ & \quad =\lim_{x\rightarrow p}\frac{f(x)+g(x)-|f(x)-g(x)|}{2}\\ & \quad =\frac{L_{f}+L_{g}-|L_{f}-L_{g}|}{2}\\ & \quad =\min\left\{L_{f},L_{g}\right\}\\ & \quad =\min\left\{\lim_{x\rightarrow p}f(x),\lim_{x\rightarrow p}g(x)\right\}.\end{align*}
The above results can be generalized to a finite number of real-valued functions. \(\sharp\)
For any \({\bf x},{\bf y}\in\mathbb{R}^{n}\), we recall
\[\parallel {\bf x}-{\bf y}\parallel=\sqrt{\sum_{i=1}^{n}\left (x_{i}-y_{i}\right )^{2}}.\]
It is clear to see
\begin{equation}{\label{ma87}}\tag{3}\left |x_{k}-y_{k}\right |\leq\parallel {\bf x}-{\bf y}\parallel\mbox{ for all }k.\end{equation}
Since
\[c_{1}^{2}+\cdots +c_{n}^{2}\leq\left (|c_{1}|+\cdots +|c_{n}|\right )^{2},\]
we also have
\begin{equation}{\label{ma88}}\tag{4}\parallel {\bf x}-{\bf y}\parallel\le\sum_{k=1}^{n}\left |x_{k}-y_{k}\right |.\end{equation}
Let \(f:S\rightarrow\mathbb{R}\) be a function defined on a subset \(S\) of \(\mathbb{R}^{n}\). Given an accumulation point \({\bf p}\) of \(S\), we write
\[\lim_{{\bf x}\rightarrow {\bf p}}f({\bf x})=b\]
when, for each \(\epsilon >0\), there exists \(\delta >0\) such that \(0<\parallel {\bf x}-{\bf p}\parallel <\delta\) implies \(|f({\bf x})-b|<\epsilon\).
Let \(f_{1},\cdots ,f_{n}\) be \(n\) real-valued functions defined on \(A\subseteq\mathbb{R}\). We define the vector-valued function \({\bf f}:A\rightarrow\mathbb{R}^{n}\) by
\[{\bf f}(x)=\left (f_{1}(x),\cdots ,f_{n}(x)\right ).\]
In this case, the functions \(f_{1},\cdots ,f_{n}\) are called the components of \({\bf f}\). We write
\[\lim_{x\rightarrow p}{\bf f}(x)={\bf a}\]
when, given any \(\epsilon >0\), there exists \(\delta >0\) such that \(|x-p|<\delta\) implies \(\parallel {\bf f}(x)-{\bf a}\parallel <\epsilon\).
Proposition B. Let \({\bf f}:A\rightarrow\mathbb{R}^{n}\) be a vector-valued function defined on a subset \(A\) of \(\mathbb{R}\). Given \({\bf a}=(a_{1},\cdots ,a_{n})\), we have
\[\lim_{x\rightarrow p}{\bf f}(x)={\bf a}\]
if and only if
\[\lim_{x\rightarrow p}f_{k}(x)=a_{k}\]
for each \(k=1,\cdots ,n\).
Proof. By referring to (\ref{ma87}), we have
\begin{align*}\left |f_{k}(x)-a_{k}\right | & \leq\parallel {\bf f}(x)-{\bf a}\parallel\\ & <\epsilon\mbox{ for all }k,\end{align*}
which implies \(|f_{k}(x)-a_{k}|<\epsilon\) for all \(k\). It follows that
\[\lim_{x\rightarrow p}{\bf f}(x)={\bf a}\]
implies
\[\lim_{x\rightarrow p}f_{k}(x)=a_{k}\]
for each \(k=1,\cdots ,n\).
For the converse, by referring to (\ref{ma88}), we have
\begin{equation}{\label{ma89}}\tag{5}\parallel {\bf f}(x)-{\bf a}\parallel\leq\sum_{k=1}^{n}\left |f_{k}(x)-a_{k}\right |. \end{equation}
Suppose that
\[\lim_{x\rightarrow p}f_{k}(x)=a_{k}\]
for each \(k=1,\cdots ,n\). Given any \(\epsilon>0\), there exists \(\delta_{k}>0\) such that \(|x-p|<\delta_{k}\) implies \(|f_{k}(x)-a_{k}|<\epsilon/n\) for \(k=1,\cdots,n\). Let
\[\delta=\min\left\{\delta_{1},\cdots,\delta_{n}\right\}.\]
If \(|x-p|<\delta\), then we have \(|x-p|<\delta_{k}\) for all \(k=1,\cdots,n\). Using (\ref{ma89}), we also have
\begin{align*} \parallel {\bf f}(x)-{\bf a}\parallel & \leq\sum_{k=1}^{n}\left |f_{k}(x)-a_{k}\right |\\ & <\sum_{k=1}^{n}\frac{\epsilon}{n}=\epsilon.\end{align*}
This shows that
\[\lim_{x\rightarrow p}f_{k}(x)=a_{k}\]
for each \(k=1,\cdots ,n\) implies
\[\lim_{x\rightarrow p}{\bf f}(x)={\bf a},\]
and the proof is complete. \(\blacksquare\)
Let \((M,d)\) be a metric space, and let \(S\) be a subset of \(M\). We consider the vector-valued functions \({\bf f}\) and \({\bf g}\) defined on \(S\) by \({\bf f}:S\rightarrow\mathbb{R}^{n}\) and \({\bf g}:S\rightarrow\mathbb{R}^{n}\). The quotients of vector-valued functions are not defined for \(n\geq 2\). The sum \({\bf f}+{\bf g}\), the scalar production \(\lambda {\bf f}\) (where \(\lambda\in\mathbb{R}\)) and the inner product \({\bf f}\bullet {\bf g}\) are defined by
\[({\bf f}+{\bf g})(x)={\bf f}(x)+{\bf g}(x),\]
\[(\lambda {\bf f})(x)=\lambda {\bf f}(x),\]
\[({\bf f}\bullet {\bf g})(x)={\bf f}(x)\bullet {\bf g}(x)\]
for each \(x\in S\). Then, we have the following rules for calculating the limits of vector-valued functions.
Proposition A. Let \(S\) be a subset of a metric space, and let \(p\) be an accumulation point of \(S\). Suppose that
\[\lim_{x\rightarrow p}{\bf f}(x)={\bf a}\]
and
\[\lim_{x\rightarrow p}{\bf g}(x)={\bf b}.\]
Then, we have the following rules.
(i) For the sum, we have
\[\lim_{x\rightarrow p}\left [{\bf f}(x)+{\bf g}(x)\right ]={\bf a}+{\bf b}.\]
(ii) For the scalar product, we have
\[\lim_{x\rightarrow p}\lambda {\bf f}(x)=\lambda {\bf a}.\]
(iii) For the inner product, we have
\[\lim_{x\rightarrow p}{\bf f}(x)\bullet {\bf g}(x)={\bf a}\bullet {\bf b}.\]
(iv) We have
\[\lim_{x\rightarrow p}\parallel {\bf f}(x)\parallel =\parallel {\bf a}\parallel .\]
Proof. We just prove parts (iii) and (iv). To prove part (iii), we have
\[{\bf f}(x)\bullet{\bf g}(x)-{\bf a}\bullet{\bf b}=
\left [{\bf f}(x)-{\bf a}\right ]\bullet\left [{\bf g}(x)-{\bf b}\right ]
+{\bf a}\bullet\left [{\bf g}(x)-{\bf b}\right ]+
{\bf b}\bullet\left [{\bf f}(x)-{\bf a}\right ].\]
Using the triangle inequality and Cauchy-Schwartz inequality, we obtain
\begin{align*} 0 & \leq\left |{\bf f}(x)\bullet{\bf g}(x)-{\bf a}\bullet{\bf b}\right |\\ & \leq\parallel {\bf f}(x)-{\bf a}\parallel\parallel {\bf g}(x)-{\bf b}\parallel +\parallel {\bf a}\parallel\parallel {\bf g}(x)-{\bf b}\parallel +\parallel {\bf b}\parallel\parallel {\bf f}(x)-{\bf a}\parallel .\end{align*}
Since each term on the right side tends to \(0\) as \(x\rightarrow p\), we have \({\bf f}(x)\bullet{\bf g}(x)\rightarrow {\bf a}\bullet{\bf b}\) as \(x\rightarrow p\).
Part (iv) follows from the inequality
\[|\parallel {\bf f}(x)\parallel-\parallel {\bf a}\parallel |\leq\parallel {\bf f}(x)-{\bf a}\parallel .\]
This completes the proof. \(\blacksquare\)
Example. We consider the vector-valued function defined by
\[{\bf f}(t)=\left (\cos (t+\pi ),\sin (t+\pi ), e^{-t^{2}}\right ).\]
Then, we have
\[\lim_{t\rightarrow 0} {\bf f}(t)=\left (\lim_{t\rightarrow 0} \cos (t+\pi ),
\lim_{t\rightarrow 0} \sin (t+\pi ),\lim_{t\rightarrow 0} e^{-t^{2}}\right )=(-1,0,1).\]
