How to find the limit of a vector function

What is the limit of a vector function?

To find the limit of a vector function,

$r(t)=a(t)\bold i+b(t)\bold j+c(t)\bold k$

we’ll need to take the limit of each term separately.

$\lim_{t\to x}[a(t)\bold i+b(t)\bold j+c(t)\bold k]$

$\lim_{t\to x}a(t)\bold i+\lim_{t\to x}b(t)\bold j+\lim_{t\to x}c(t)\bold k$

$a(x)\bold i+b(x)\bold j+c(x)\bold k$

Example

Find the limit of the vector function.

$\lim_{t\to0}\left[(t^2-2)\bold i+\ln{(t+e)}\bold j+\frac{4t}{\sin{t}}\bold k\right]$

We’ll take the limit of each term separately.

$\lim_{t\to0}(t^2-2)\bold i+\lim_{t\to0}\ln{(t+e)}\bold j+\lim_{t\to0}\frac{4t}{\sin{t}}\bold k$

Evaluating the first two terms as $t\to0$ , we get

$(0^2-2)\bold i+\ln{(0+e)}\bold j+\lim_{t\to0}\frac{4t}{\sin{t}}\bold k$

$-2\bold i+\ln{(e)}\bold j+\lim_{t\to0}\frac{4t}{\sin{t}}\bold k$

$-2\bold i+1\bold j+\lim_{t\to0}\frac{4t}{\sin{t}}\bold k$

$-2\bold i+\bold j+\lim_{t\to0}\frac{4t}{\sin{t}}\bold k$

Because the third term gives $0/0$ when $t\to0$ , we have to use L’Hospital’s rule, replacing the numerator and denominator with their derivatives.

$-2\bold i+\bold j+\lim_{t\to0}\frac{4}{\cos{t}}\bold k$

Evaluating as $t\to0$ , we get

$-2\bold i+\bold j+\frac{4}{\cos{0}}\bold k$

$-2\bold i+\bold j+\frac41\bold k$

$-2\bold i+\bold j+4\bold k$

This is the limit of the vector function.