Lim x - 027-x1-3-3-9-27-x3 equals-

The correct option is A 16lim _x → 0(27+x)^1/3 3/9 (27+x)^2/3 By using L'Hopitals rule, as it is in the form 0/0 nbsp; we get lim _x → 0, [1/2(27+x)^ 2/3/ ...

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Standard XII

Mathematics

Standard Limits to Remove Indeterminate Form

lim x → 027+x...

Question

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

- \frac{1}{6}

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\frac{1}{6}

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\frac{1}{3}

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- \frac{1}{3}

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Solution

The correct option is A $- \frac{1}{6}$
$lim x \to 0 \frac{(27 + x)^{\frac{1}{3}} - 3}{9 - (27 + x)^{\frac{2}{3}}}$
By using L'Hopitals rule, as it is in the form $\frac{0}{0}$ we get-
$lim x \to 0, - ⎡ ⎣ \frac{1}{2} \frac{(27 + x)^{\frac{- 2}{3}}}{(27 + x)^{\frac{- 1}{3}}} ⎤ ⎦ = - \frac{1}{6}$

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Similar questions

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

equals.

lim x \to 0 \frac{(27 + x)^{\frac{1}{3}} - 3}{9 - (27 + x)^{\frac{2}{3}}} =

{lim}_{x \to 0} \frac{(27 + x)^{1 / 3} - 3}{9 - (27 + x)^{2 / 3}}

equals :

lim x \to 3 \frac{x^{3} - 7 x^{2} + 15 x - 9}{x^{4} - 5 x^{3} + 27 x - 27}

is equal to

Q. Solve:

lim x \to 27 \frac{(x^{1 / 3} + 3) (x^{1 / 3} - 3)}{x - 27}

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Standard Limits to Remove Indeterminate Form

Standard XII Mathematics

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limx→0(27+x)13−39−(27+x)23 equals :

The correct option is A −16limx→0(27+x)13−39−(27+x)23 By using L'Hopitals rule, as it is in the form 00 we get- limx→0,−⎡⎣12(27+x)−23(27+x)−13⎤⎦=−16

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