Limits to the Limit -2!

Calculus Level 3

$\large \lim_{x \to 0 } \left ( \frac{\sqrt{1+ \tan (x)} - \sqrt{1 + \sin(x)}}{x^3} \right)$

Find the reciprocal of the limit above.

The answer is 4.

1 solution

Muhammad Fairuzi Teguh
Jun 5, 2015

$\lim _{ x\rightarrow 0 }{ \left( \frac { \sqrt { 1+tanx } -\sqrt { 1+sinx } }{ { x }^{ 3 } } \right) = } \lim _{ x\rightarrow 0 }{ \left( \frac { tanx+sinx }{ { x }^{ 3 }(\sqrt { 1+tanx } +\sqrt { 1+sinx } ) } \right) = } \lim _{ x\rightarrow 0 }{ \left( \frac { sinx(1+cosx) }{ { x }^{ 3 }(\sqrt { 1+tanx } +\sqrt { 1+sinx } )cosx } \right) }$

$=\lim _{ x\rightarrow 0 }{ \left( \frac { sinx(1-\cos ^{ 2 }{ x } ) }{ { x }^{ 3 }(\sqrt { 1+tanx } +\sqrt { 1+sinx } )cosx(1-cosx) } \right) } =\lim _{ x\rightarrow 0 }{ \left( \frac { \sin ^{ 3 }{ x } }{ { x }^{ 3 }(\sqrt { 1+tanx } +\sqrt { 1+sinx } )cosx(1-cosx) } \right) }$

$=\lim _{ x\rightarrow 0 }{ \left( \frac { \frac { \sin ^{ 3 }{ x } }{ { x }^{ 3 } } }{ (\sqrt { 1+tanx } +\sqrt { 1+sinx } )cosx(1-cosx) } \right) } =\frac { 1 }{ 4 }$

The answer is $4$

Wrong solution. The numerator becomes $\tan x - \sin x$ . You were lucky to get the correct answer

Krutarth Patel - 1 month ago

Limits to the Limit -2!

The answer is 4.

1 solution

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