What is the limit as x approaches 0 of ((1+x)^9-1)/x ?

The limit as x approaches 0 of ((1+x)^9-1)/x is 9

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limit as x approaches 0 of ((1+x)^9-1)/x

x \to 0 lim (\frac{( 1 + x ) ^{9} - 1}{x})

9

Solution steps

x \to 0 lim (\frac{( 1 + x ) ^{9} - 1}{x})

Simplify

\frac{( 1 + x ) ^{9} - 1}{x} : (x^{2} + 3 x + 3) ((x + 1)^{6} + (x + 1)^{3} + 1)

= x \to 0 lim ((x^{2} + 3 x + 3) ((x + 1)^{6} + (x + 1)^{3} + 1))

Plug in the value

x = 0

= (0^{2} + 3 \cdot 0 + 3) ((0 + 1)^{6} + (0 + 1)^{3} + 1)

Simplify

(0^{2} + 3 \cdot 0 + 3) ((0 + 1)^{6} + (0 + 1)^{3} + 1) : 9

= 9

\frac{\partial}{\partial x} (\frac{42}{( 5 + x ^{2} + y ^{2} )})

\int \frac{1}{1 + cos ( 2 x )} d x

\int x (x^{2} + 1)^{5} d x

x = 3 y^{2}, x = 4 + 2 y^{2}

x \to - 4 lim (x^{2} - 3)

What is the limit as x approaches 0 of ((1+x)^9-1)/x ?
The limit as x approaches 0 of ((1+x)^9-1)/x is 9