Limits of Vector Functions

Definition: If ${\vec {r}}(t)=(x(t),y(t),z(t))$ is a vector-valued function, then $\lim _{t\to a}{\vec {r}}(t)=\left(\lim _{t\to a}x(t),\lim _{t\to a}y(t),\lim _{t\to a}z(t)\right)$ provided that the limits of the components exist.

//Limits of vector-valued functions in $\mathbb {R} ^{n}$ are defined similarly as the limit of each component. //

Let's look at some examples of evaluating limits of vector-valued functions. Consider the vector-valued function ${\vec {r}}(t)=(t^{2}-1,t+1,e^{t})$ and suppose that we wanted to compute $\lim _{t\to 2}{\vec {r}}(t)$. To compute this limit, all we need to do is compute the limits of the components.

${\begin{aligned}\quad \lim _{t\to 2}{\vec {r}}(t)=\left(\lim _{t\to 2}t^{2}-1,\lim _{t\to 2}t+1,\lim _{t\to 2}e^{t}\right)=(3,3,e^{2})\end{aligned}}$

For another example, consider the vector-valued function ${\vec {r}}(t)=\left({\frac {e^{t}-1}{t}},{\frac {t-1}{t+1}},t^{2}+3\right)$ and suppose that we wanted to compute $\lim _{t\to 0}{\vec {r}}(t)$. To compute this limit, we will compute all of the limits of the components again, however, this time the limits are a little trickier to compute. Fortunately, we have already learned about various rules to evaluate limits.

${\begin{aligned}\quad \lim _{t\to 0}{\vec {r}}(t)=\left(\lim _{t\to 0}{\frac {e^{t}-1}{t}},\lim _{t\to 0}{\frac {t-1}{t+1}},\lim _{t\to 0}t^{2}+3\right)\end{aligned}}$

For $\lim _{t\to 0}{\frac {e^{t}-1}{t}}$ we will use L'Hospital's Rule, and so $\lim _{t\to 0}{\frac {e^{t}-1}{t}}{\overset {H}{=}}\lim _{t\to 0}{\frac {e^{t}}{1}}=1$.

For $\lim _{t\to 0}{\frac {t-1}{t+1}}$, we can use direct substitution and so $\lim _{t\to 0}{\frac {t-1}{t+1}}=-1$.

Now $\lim _{t\to 0}t^{2}+3$ is also easy to compute by direct substitution and so $\lim _{t\to 0}t^{2}+3=3$.

Thus we have that $\lim _{t\to 0}{\vec {r}}(t)=(1,-1,3)$.

The following theorem gives us a formal definition to say a vector-valued function ${\vec {r}}(t)$ has limit ${\vec {b}}$ at $t=a$, which is analogous to that of limits of real-valued functions.

Theorem 1: Let ${\vec {r}}(t)=(x(t),y(t),z(t))$ be a vector-valued function and let ${\vec {b}}=(b_{1},b_{2},b_{3})\in \mathbb {R} ^{3}$. Then $\lim _{t\to a}{\vec {r}}(t)={\vec {b}}$ if and only if $\forall \epsilon >0$ $\exists \delta >0$ such that if $0<\mid t-a\mid <\delta$ then $\|{\vec {r}}(t)-{\vec {b}}\|<\epsilon$.

• Proof: $\Rightarrow$ Suppose that $\lim _{t\to a}{\vec {r}}(t)={\vec {b}}$. Then we have that:

${\begin{aligned}\left(\lim _{t\to a}x(t),\lim _{t\to a}y(t),\lim _{t\to a}z(t)\right)=(b_{1},b_{2},b_{3})\end{aligned}}$

• Now recall that two vectors are equal if and only if their components are equal, and so the equation above implies that $\lim _{t\to a}x(t)=b_{1}$, $\lim _{t\to a}y(t)=b_{2}$, and $\lim _{t\to a}z(t)=b_{3}$. Now notice that these three limits are limits of real-valued functions.

• Since $\lim _{t\to a}x(t)=b_{1}$ then $\forall \epsilon >0$ $\exists \delta _{1}>0$ such that if $0<\mid t-a\mid <\delta _{1}$ then $\mid x(t)-b_{1}\mid <{\frac {\epsilon }{3}}$.

• Since $\lim _{t\to a}y(t)=b_{2}$ then $\forall \epsilon >0$ $\exists \delta _{2}>0$ such that if $0<\mid t-a\mid <\delta _{2}$ then $\mid y(t)-b_{2}\mid <{\frac {\epsilon }{3}}$.

• Since $\lim _{t\to a}z(t)=b_{3}$ then $\forall \epsilon >0$ $\exists \delta _{3}>0$ such that if $0<\mid t-a\mid <\delta _{3}$ then $\mid z(t)-b_{3}\mid <{\frac {\epsilon }{3}}$.

• Let $\delta =\mathrm {min} \{\delta _{1},\delta _{2},\delta _{3}\}$. Then if $0<\mid t-a\mid <\delta$ we have that:

${\begin{aligned}\quad \quad \|{\vec {r}}(t)-{\vec {b}}\|=\|(x(t)-b_{1},y(t)-b_{2},z(t)-b_{3}))\|={\sqrt {(x(t)-b_{1})^{2}+(y(t)-b_{2})^{2}+(z(t)-b_{3})^{2}}}\\\quad \quad \leq {\sqrt {(x(t)-b_{1})^{2}}}+{\sqrt {(y(t)-b_{2})^{2}}}+{\sqrt {(z(t)-b_{3})^{2}}}=\mid x(t)-b_{1}\mid +\mid y(t)-b_{2}\mid +\mid z(t)-b_{3}\mid <{\frac {\epsilon }{3}}+{\frac {\epsilon }{3}}+{\frac {\epsilon }{3}}=\epsilon \end{aligned}}$

• $\Leftarrow$ Suppose that $\forall \epsilon >0$ $\exists \delta >0$ such that if $0<\mid t-a\mid <\delta$ then $\|{\vec {r}}(t)-{\vec {b}}\|<\epsilon$. Therefore we have that

$\|{\vec {r}}(t)-{\vec {b}}\|=\|(x(t)-b_{1},y(t)-b_{2},z(t)-b_{3})\|<\epsilon$, which implies that:

${\begin{aligned}{\sqrt {(x(t)-b_{1})^{2}+(y(t)-b_{2})^{2}+(z(t)-b_{3})^{2}}}<\epsilon \\(x(t)-b_{1})^{2}+(y(t)-b_{2})^{2}+(z(t)-b_{3})^{2}<\epsilon ^{2}\\\end{aligned}}$

• Now since all terms of the lefthand side of this equation are positive, we must have that for $0<\mid t-a\mid <\delta$ then $(x(t)-b_{1})^{2}<\epsilon ^{2}$, $(y(t)-b_{2})^{2}<\epsilon ^{2}$, and $(z(t)-b_{3})^{2}<\epsilon ^{2}$, and so $\mid x(t)-b_{1}\mid <\epsilon$, $\mid y(t)-b_{2}\mid <\epsilon$ and $\mid z(t)-b_{3}\mid <\epsilon$. Therefore by the definition of real-valued function limits we have that $\lim _{t\to a}x(t)=b_{1}$, $\lim _{t\to a}y(t)=b_{2}$, and $\lim _{t\to a}z(t)=b_{3}$.

• Thus $\lim _{t\to a}{\vec {r}}(t)=\left(\lim _{t\to a}x(t),\lim _{t\to a}y(t),\lim _{t\to a}z(t)\right)=(b_{1},b_{2},b_{3})={\vec {b}}$. $\blacksquare$

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Content obtained and/or adapted from:

• Limits of Vector-Valued Functions, mathonline.wikidot.com under a CC BY-SA license