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

Given:

f(x) = (x − a₁)(x − a₂) … (x − a_n)

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Part (i): Evaluate \( \lim_{x \to a_1} f(x) \)

As x → a₁, the factor (x − a₁) → 0.

All other factors (x − a₂), … , (x − a_n) approach finite non-zero real numbers.

Therefore,

f(x) → 0 × (a₁ − a₂) … (a₁ − a_n) = 0

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Result:

\( \lim_{x \to a_1} f(x) = \) 0

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Part (ii): Evaluate \( \lim_{x \to a} f(x) \), where a ≠ a₁, a₂, … , a_n

Since a is different from all a₁, a₂, … , a_n, none of the factors become zero.

The function f(x) is a polynomial and hence continuous for all real x.

Therefore,

\( \lim_{x \to a} f(x) = f(a) \)

= (a − a₁)(a − a₂) … (a − a_n)

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Final Answers:

\( \lim_{x \to a_1} f(x) = \) 0

\( \lim_{x \to a} f(x) = \) (a − a₁)(a − a₂) … (a − a_n)