Limx-x7-1-x5-2-x9-2-x6-1-x5-1-x4-4-

The correct option is A 1 lim_x→∞√(x^7 1)√(x^5+2)√(x^9 2)/√(x^6+1)√(x^5+1)√(x^4+4) =lim_x→∞x^3(1 1/x^7)^1/7(1+2/x^5)^1/5(1 2/x^9)^1/9/x^3(1+1/x^ ...

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Byju's Answer

Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

limx→∞√x7-1√x...

Question

$lim x \to \infty \frac{\sqrt[7]{x^{7} - 1} \sqrt[5]{x^{5} + 2} \sqrt[9]{x^{9} - 2}}{\sqrt[6]{x^{6} + 1} \sqrt[5]{x^{5} + 1} \sqrt[4]{x^{4} + 4}} =$

1

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Solution

The correct option is A $1$
$lim x \to \infty \frac{\sqrt[7]{x^{7} - 1} \sqrt[5]{x^{5} + 2} \sqrt[9]{x^{9} - 2}}{\sqrt[6]{x^{6} + 1} \sqrt[5]{x^{5} + 1} \sqrt[4]{x^{4} + 4}}$
$= lim x \to \infty \frac{x^{3} (1 - \frac{1}{x^{7}})^{\frac{1}{7}} (1 + \frac{2}{x^{5}})^{\frac{1}{5}} (1 - \frac{2}{x^{9}})^{\frac{1}{9}}}{x^{3} (1 + \frac{1}{x^{6}})^{\frac{1}{6}} (1 + \frac{1}{x^{5}})^{\frac{1}{5}} (1 + \frac{4}{x^{4}})^{\frac{1}{4}}}$
$= 1$

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Similar questions

You are given $c o s x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} . . . . . .;$

$s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} . . . . . .$

$t a n x = x + \frac{x^{3}}{3} + \frac{2. x^{5}}{15} . . . . . .$

Find the value of $lim x \to 0 \frac{x c o s x + s i n x}{x^{2} + t a n x}$

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You are given $c o s x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} . . . . . .;$

$s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} . . . . . .$

$t a n x = x + \frac{x^{3}}{3} + \frac{2. x^{5}}{15} . . . . . .$

Find the value of $lim x \to 0 \frac{x c o s x + s i n x}{x^{2} + t a n x}$

$a = 1 + \frac{x^{3}}{3!} + \frac{x^{6}}{6!} + . . .$ ,

$b = x + \frac{x^{4}}{4!} + \frac{x^{7}}{7!} + . . .$ ,

$c = \frac{x^{2}}{2!} + \frac{x^{5}}{5!} + \frac{x^{8}}{8!} + . . .$ .

If $a^{3} + b^{3} + c^{3} - z a b c = 1$ . Find the value of $z$

Q. If the series

1 + \frac{x^{3}}{⌊ 3} + \frac{x^{6}}{⌊ 6} + . . . ., x + \frac{x^{4}}{⌊ 4} + \frac{x^{7}}{⌊ 7} + . . . . ., \frac{x^{2}}{⌊ 2} + \frac{x^{5}}{⌊ 5} + \frac{x^{8}}{⌊ 8} + . . . . .

are denoted by a, b, c respectively, show that

a^{3} + b^{3} + c^{3} - 3 a b c = 1

You are given
$cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} . . . . . .$ ;
$sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} . . . . . .$ ;
$tan x = x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} . . . . . .$
Then the value of
$lim x \to 0 \frac{x cos x + sin x}{x^{2} + tan x}$ is

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limx→∞7√x7−15√x5+29√x9−26√x6+15√x5+14√x4+4=

The correct option is A 1limx→∞7√x7−15√x5+29√x9−26√x6+15√x5+14√x4+4=limx→∞x3(1−1x7)17(1+2x5)15(1−2x9)19x3(1+1x6)16(1+1x5)15(1+4x4)14=1

$lim x \to \infty \frac{\sqrt[7]{x^{7} - 1} \sqrt[5]{x^{5} + 2} \sqrt[9]{x^{9} - 2}}{\sqrt[6]{x^{6} + 1} \sqrt[5]{x^{5} + 1} \sqrt[4]{x^{4} + 4}} =$