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Question

The value of $$\displaystyle \lim_{x\rightarrow 0}\frac{\sqrt{x^{2}+1}-1}{\sqrt{x^{2}+16}-4}$$ is


A
3
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B
4
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C
1
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D
2
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Solution

The correct option is A 4
$$\displaystyle \lim _{ x\rightarrow 0 } \dfrac { \sqrt { x^{ 2 }+1 } -1 }{ \sqrt { x^{ 2 }+16 } -4 } \\ =\displaystyle \lim _{ x\rightarrow 0 } \dfrac { \sqrt { x^{ 2 }+1 } -1 }{ \sqrt { x^{ 2 }+16 } -4 } .\dfrac { \sqrt { x^{ 2 }+1 } +1 }{ \sqrt { x^{ 2 }+1 } +1 } .\dfrac { \sqrt { x^{ 2 }+16 } +4 }{ \sqrt { x^{ 2 }+16 } +4 } \\ =\displaystyle \lim _{ x\rightarrow 0 } \dfrac { \left( { x }^{ 2 } \right) \left( \sqrt { x^{ 2 }+16 } +4 \right)  }{ \left( { x }^{ 2 } \right) \left( \sqrt { x^{ 2 }+1 } +1 \right)  } =\displaystyle \lim _{ x\rightarrow 0 } \dfrac { \left( \sqrt { x^{ 2 }+16 } +4 \right)  }{ \left( \sqrt { x^{ 2 }+1 } +1 \right)  } =\cfrac { 8 }{ 2 } =4$$
Hence, option 'B' is correct.

Mathematics

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