Question

Evaluate: Lt x 0 sqrt 1+x -1 sqrt[3] 1+x -1 .​

Answer 1

Answer:

The answer is $\dfrac{3}{2}$

Step-by-step explanation:

Given limit is

$=\displaystyle \lim_{x\to 0}\dfrac{\sqrt{1+x}-1}{\sqrt[3]{1+x}-1}$

Substitute $1+x=y^6$ $\Rightarrow x\to 0\Rightarrow y\to 1$

$=\displaystyle \lim_{y\to 1}\dfrac{\sqrt{y^6}-1}{\sqrt[3]{y^6}-1}\\\\=\displaystyle \lim_{y\to 1}\dfrac{y^3-1}{y^2-1}=\displaystyle \lim_{y\to 1}\dfrac{(y-1)(y^2+y+1)}{(y-1)(y+1)}\\\\=\displaystyle \lim_{y\to 1} \dfrac{y^2+y+1}{y+1}=\dfrac{1^2+1+1}{1+1}=\dfrac{3}{2}$

      Identities used :

1. $a^2-b^2=(a-b)(a+b)$
2. $a^3-b^3=(a-b)(a^2+ab+b^2)$