Question

Lim. 8x³-1 / 16x4-1
x--->1/2

Answer 1

Answer:

$\bf\lim_{x\to\frac{1}{2}}\;\frac{8x^3-1}{16x^4-1}=\frac{3}{4}$

Step-by-step explanation:

Given:

$\lim_{x\to\frac{1}{2}}\;\frac{8x^3-1}{16x^4-1}$

we factorize both numerator and denominator

$=\lim_{x\to\frac{1}{2}}\;\frac{(2x)^3-1^3}{(4x^2)^2-1^2}$

Using

$\boxed{\bf\;a^3-b^3=(a-b)(a^2+ab+b^2)}$

$\boxed{\bf\;a^2-b^2=(a-b)(a+b)}$

$=\lim_{x\to\frac{1}{2}}\;\frac{(2x-1)(4x^2+2x+1)}{(4x^2-1)(4x^2+1)}$

$=\lim_{x\to\frac{1}{2}}\;\frac{(2x-1)(4x^2+2x+1)}{((2x)^2-1^2)(4x^2+1)}$

$=\lim_{x\to\frac{1}{2}}\;\frac{(2x-1)(4x^2+2x+1)}{(2x-1)(2x+1)(4x^2+1)}$

$=\lim_{x\to\frac{1}{2}}\;\frac{4x^2+2x+1}{(2x+1)(4x^2+1)}$

$=\frac{4(\frac{1}{4})+2(\frac{1}{2})+1}{(2(\frac{1}{2})+1)(4(\frac{1}{4})+1)}$

$=\frac{1+1+1}{1+1)(1+1)}$

$=\frac{3}{4}$

$\implies\boxed{\bf\lim_{x\to\frac{1}{2}}\;\frac{8x^3-1}{16x^4-1}=\frac{3}{4}}$