Evaluate- lim x- 0 - 1-x 2 - 1-x - 1- x 3 - 1-x

x→ 0lt√(1+x^2) √(1 x)√(1+x^3) √(1+x) which is of the form 00 Hence, Applying L'Hospital Rule, we get x→ 0Lt2x2√(1+x^2) 12√(1+x)3x^22√(1+x^3) 12√(1 ...

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Byju's Answer

Standard XII

Mathematics

Integration Using Substitution

Evaluate: l...

Question

Evaluate: ${lim}_{x \to 0} \frac{\sqrt{{1 + x}^{2}} - \sqrt{1 + x}}{\sqrt{1 + x^{3}} - \sqrt{1 + x}}$

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Solution

$l t x \to 0 \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x}}{\sqrt{1 + x^{3}} - \sqrt{1 + x}}$
which is of the form $\frac{0}{0}$
Hence, Applying L'Hospital Rule, we get
$L t x \to 0 \frac{\frac{2 x}{2 \sqrt{1 + x^{2}}} - \frac{1}{2 \sqrt{1 + x}}}{\frac{3 x^{2}}{2 \sqrt{1 + x^{3}}} - \frac{1}{2 \sqrt{1 + x}}}$
$L t x \to 0 \frac{2 x \sqrt{1 + x} - \sqrt{1 + x^{2}}}{3 x^{2} \sqrt{1 + x} - \sqrt{1 + x^{3}}} \cdot (\frac{\sqrt{1 + x^{3}}}{\sqrt{1 + x^{2}}})$
$= (\frac{0 - \sqrt{1}}{0 - \sqrt{1}}) (\frac{\sqrt{1}}{\sqrt{1}})$
$= 1 (1) = 1$ .

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Evaluate: limx→0√1+x2−√1+x√1+x3−√1+x

ltx→0√1+x2−√1−x√1+x3−√1+xwhich is of the form 00Hence, Applying L'Hospital Rule, we getLtx→02x2√1+x2−12√1+x3x22√1+x3−12√1+xLtx→02x√1+x−√1+x23x2√1+x−√1+x3⋅(√1+x3√1+x2)=(0−√10−√1)(√1√1)=1(1)=1.

Evaluate: ${lim}_{x \to 0} \frac{\sqrt{{1 + x}^{2}} - \sqrt{1 + x}}{\sqrt{1 + x^{3}} - \sqrt{1 + x}}$