If \(\lim\limits_{x \to a}\frac{x^9-a^9}{x-a}\) = \(\lim\limits_{x \

1.	If \(\lim\limits_{x \to a}\frac{x^9-a^9}{x-a}\) = \(\lim\limits_{x \to 3}(x+6)\), find the value of a.
Answer» \(\lim\limits_{x \to a}\frac{x^9-a^9}{x-a}\) = \(\lim\limits_{x \to 3}(x+6)\) 9.a^{9 - 1} = 3 + 6 9.a⁸ = 9 a⁸ = 1 Taking squareroot on bothsides, we get \((a^8)^{\frac{1}{2}}\)= ±1 a⁴ = ±1 But a⁴= -1 is imposssible. ∴ a⁴ = 1 Again taking squareroot, we get \((a^4)^{\frac{1}{2}}\)= ±1 a² = ±1 a² = -1 is imposssible ∴ a² = 1 Again taking positive squareroot, a = ±1

If \(\lim\limits_{x \to a}\frac{x^9-a^9}{x-a}\) = \(\lim\limits_{x \to 3}(x+6)\), find the value of a.

Answer»

\(\lim\limits_{x \to a}\frac{x^9-a^9}{x-a}\) = \(\lim\limits_{x \to 3}(x+6)\)

9.a^{9 - 1} = 3 + 6

9.a⁸ = 9

a⁸ = 1

Taking squareroot on bothsides, we get

\((a^8)^{\frac{1}{2}}\)= ±1

a⁴ = ±1

But a⁴= -1 is imposssible.

∴ a⁴ = 1

Again taking squareroot, we get

\((a^4)^{\frac{1}{2}}\)= ±1

a² = ±1

a² = -1 is imposssible

∴ a² = 1

Again taking positive squareroot, a = ±1