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.