Question

Let a₁, a₂,..., a_n be fixed real numbers and define a function f(x)=(x-a₁)(x-a₂)...(x-a_n).
What is \(\lim_{x\rightarrow a_1}\) f(x)? 
For some a ≠ a₁, a₂ .....a_n, Compute \(\lim_{x\rightarrow a}\) f(x).

Answer 1

The given function is f(x) = (x − a₁) ( x − a₂)... ( x − a_n)
\(\lim_{x\rightarrow a_1}\) f(x)= \(\lim_{x\rightarrow a_1}\) [(x-a₁)(x − a₂).....(x − a_n)]
=[\(\lim_{x\rightarrow a_1}\) (x − a₁) ][\(\lim_{x\rightarrow a_1}\)(x − a₂)]..... [\(\lim_{x\rightarrow a_1}\) (x-a_n)]
=(a₁-a₁)(a₁-a₂)....(a₁ -a_n) = 0
∴\(\lim_{x\rightarrow a_1}\) f(x)=0
Now, \(\lim_{x\rightarrow a}\) f(x)= \(\lim_{x\rightarrow a}\)[(x − a₁)(x-a₂)...(x-a_n)]
= [\(\lim_{x\rightarrow (x-a_1)}\)][\(\lim_{x\rightarrow a}\) (x-a₂)]...\(\lim_{x\rightarrow a}\)(x-a_n)]
=(a-a₁) (a-a₂)....(a-a_n)
∴\(\lim_{x\rightarrow a}\) f(x)=(a-a₁)(a-a₂)...(a-a_n)