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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???

How to find the limit of a vector function


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Finding the limit of the vector function

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.


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