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Limits Continuity and Differentiability question

2018 · 16 Apr · Shift 1 · Q32

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JEE Main
Mathematics
Limits Continuity and Differentiability
2018 · 16 Apr · Shift 1 · Q32

Limits Continuity and Differentiability question

2018 · 16 Apr · Shift 1 · Q32

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

\mathop {\lim }\limits_{x \to 0} \,\,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}

equals.

A
${1 \over 3}$
B
$-{1 \over 3}$
C
$-{1 \over 6}$
D
${1 \over 6}$

View written solutionFree

Correct answer: C

Let $t=(27+x)^{1/3}.$ Then as $x\to 0$ , we have $t\to 3$ .

The given limit becomes

\lim_{t\to 3}\frac{t-3}{9-t^2}.$$ 2. Factor the denominator: $$9-t^2=(3-t)(3+t)=-(t-3)(t+3).$$ So, $$\frac{t-3}{9-t^2}= rac{t-3}{-(t-3)(t+3)}=-\frac{1}{t+3}, \quad t\ne 3.$$ 3. Now take the limit: $$\lim_{t\to 3}-\frac{1}{t+3}=-\frac{1}{3+3}=-\frac{1}{6}.$$ 4. Therefore, the correct option is $$\boxed{\text{C: }-\frac{1}{6}}.$$

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