Lim x- 1 - 7- x 3 - 3- x 2 - x-1

The given Expression is of the form, 00 , so apply L'hospital's #160;rule.Differentiating the numerator, we get #160; x^2(7+x^3)^ 2/3 x.(3+x^2)^ 1/ ...

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Standard IX

Mathematics

Polynomial and Its General Form

lim x→ 1 √ 7+...

Question

$lim x \to 1 \frac{\sqrt[3]{7 + x^{3}} - \sqrt{3 + x^{2}}}{x - 1}$

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Solution

The given Expression is of the form, $\frac{0}{0}$ , so apply L'hospital's rule.
Differentiating the numerator, we get
$x^{2} (7 + x^{3})^{- 2 / 3} - x . (3 + x^{2})^{- 1 / 2}$
Differentiating the denominator we get 1.
$x \to 1$
Substituting, we get $8^{- 2 / 3} - 4^{- 1 / 2} = - \frac{1}{4}$

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lim x \to 1 \frac{\sqrt[3]{7 + x^{3}} - \sqrt{3 + x^{2}}}{x - 1}

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Evaluate the following one sided limits:

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$(i i i) {lim}_{x \to 0^{+}} \frac{1}{3 x}$

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$(v i) {lim}_{x \to \frac{π^{-}}{2}} t a n x$

$(v i i) {lim}_{x \to \frac{π}{2^{+}}} s e c x$

$(v i i i) {lim}_{x \to 0^{-}} \frac{x^{2} - 3 x + 2}{x^{3} - 2 x^{2}}$

$(i x) {lim}_{x \to - 2^{+}} \frac{x^{2} - 1}{2 x + 4}$

$(x) {lim}_{x \to 0^{+}} (2 - c o t x)$

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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

___

Q. Solve:

lim x \to \infty \frac{\sqrt{x^{2} + 1} - 3 \sqrt{x^{3} + 1}}{4 \sqrt{x^{4} + 1} + 5 \sqrt{x^{4} + 1}}

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limx→13√7+x3−√3+x2x−1

The given Expression is of the form,00 , so apply L'hospital's rule.Differentiating the numerator, we get x2(7+x3)−2/3−x.(3+x2)−1/2Differentiating the denominator we get 1.x→1Substituting, we get 8−2/3−4−1/2=−14

$lim x \to 1 \frac{\sqrt[3]{7 + x^{3}} - \sqrt{3 + x^{2}}}{x - 1}$

The given Expression is of the form, $\frac{0}{0}$ , so apply L'hospital's rule.
Differentiating the numerator, we get
$x^{2} (7 + x^{3})^{- 2 / 3} - x . (3 + x^{2})^{- 1 / 2}$
Differentiating the denominator we get 1.
$x \to 1$
Substituting, we get $8^{- 2 / 3} - 4^{- 1 / 2} = - \frac{1}{4}$