Question

limit X tends to 2 x3-8/x2-4​

Answer 1

Given to find,

$\displaystyle\longrightarrow L=\lim_{x\to2}\dfrac{x^3-8}{x^2-4}$

or,

$\displaystyle\longrightarrow L=\lim_{x\to2}\dfrac{x^3-2^3}{x^2-2^2}$

We know that,

• $a^3-b^3=(a-b)(a^2+ab+b^2)$

• $a^2-b^2=(a-b)(a+b)$

Thus, factorising numerator and denominator each we get,

$\displaystyle\longrightarrow L=\lim_{x\to2}\dfrac{(x-2)(x^2+2x+2^2)}{(x-2)(x+2)}$

$\displaystyle\longrightarrow L=\lim_{x\to2}\dfrac{x^2+2x+4}{x+2}$

Putting $x=2,$

$\displaystyle\longrightarrow L=\dfrac{2^2+2\times2+4}{2+2}$

$\displaystyle\longrightarrow L=\dfrac{12}{4}$

$\displaystyle\longrightarrow\underline{\underline{L=3}}$

Hence 3 is the answer.

Answer 2

Answer:

3

Step-by-step explanation:

LIM x tends to 3

$\frac{ {x}^{3} - 8 } { {x}^{2} - 4}$

a³-b³ = (a-b) (a²ab+b²)

LIM X tends to 3

$\frac{(x - 2)( {x}^{2} + 2x + 4)}{(x - 2)(x + 2)}$

LIM x tends to 3

$\frac{( {x}^{2} + 2x + 4)}{(x + 2)}$

Now put x as 3

We get

$\frac{4 + 4 + 4}{4}$

Ans = 3