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Question

Evaluate
$$lim_{x\to 0} \dfrac{\sqrt[K] {1+x} -1}{x}$$ ( K is a positive integer )


A
K
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B
-K
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C
1K
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D
1K
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Solution

The correct option is B $$\dfrac{1}{K}$$
$$\underset { x\rightarrow 0 }{ lim } $$  $$\dfrac { \sqrt [ K ]{ 1+x } -1 }{ x } $$        ($$\frac { 0 }{ 0 } $$ form)

$$=\underset { x\rightarrow 0 }{ lt } $$  $$\dfrac { \dfrac { 1 }{ K } { \left( 1+x \right)  }^{ K-1 } }{ 1 } $$        (Using $$L$$ Hospitals rule)

$$=\dfrac { 1 }{ K } { \left( 1+0 \right)  }^{ K-1 }$$

$$=\dfrac { 1 }{ K } { \left( 1 \right)  }^{ K-1 }$$

$$=\dfrac { 1 }{ K } $$

Mathematics

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