limx→0((27+x)^13-3)/(9-(27+x)^23) equals

Correct answer: (A) -1/6. Given limit is limx→0(27+x^13-3 )/(9-27+x^23) 0/0 form Applying L'Hospital's rule, ⇒limx→0 (1327+x^(-23) )/(-2327+x^(-13))=27^(-23)-227^(-13) =(-1)/(227^13 )=-1/6.

Q38 / 90 · JEE Main 2018 (16 Apr Online) · Mathematics

$\lim_{x \to 0} \frac{{(27 + x)}^{\frac{1}{3}} - 3}{9 - {(27 + x)}^{\frac{2}{3}}}$ equals

Correct answer
A
$- \frac{1}{6}$
B
$\frac{1}{6}$
C
$\frac{1}{3}$
D
$- \frac{1}{3}$

Solution

Given limit is

\lim_{x \to 0} \frac{{(27 + x)}^{\frac{1}{3}} - 3}{9 - {(27 + x)}^{\frac{2}{3}}} (\frac{0}{0} f o r m)

Applying L'Hospital's rule,

\Rightarrow \underset{x \to 0}{l i m} \frac{\frac{1}{3} {(27 + x)}^{- \frac{2}{3}}}{- \frac{2}{3} {(27 + x)}^{- \frac{1}{3}}} = \frac{{(27)}^{- \frac{2}{3}}}{- 2 {(27)}^{- \frac{1}{3}}}

= \frac{- 1}{2 {(27)}^{\frac{1}{3}}} = - \frac{1}{6}

Hence, the correct answer is option: 'A'.

limx→0(27+x)13-39-(27+x)23 equals

Solution

Relevant Questions

$\lim_{x \to 0} \frac{{(27 + x)}^{\frac{1}{3}} - 3}{9 - {(27 + x)}^{\frac{2}{3}}}$ equals