What is the limit as x approaches 0 of (1+x)^2-1 ?

The limit as x approaches 0 of (1+x)^2-1 is 0

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limit as x approaches 0 of (1+x)^2-1

x \to 0 lim ((1 + x)^{2} - 1)

0

Solution steps

x \to 0 lim ((1 + x)^{2} - 1)

Plug in the value

x = 0

= (1 + 0)^{2} - 1

Simplify

= 0

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

\int \frac{1}{4 + v ^{2}} d v

f (x) = \frac{1 - x ^{2}}{x}

y = 10 csc (5 x^{5} - 4 x + 4)

What is the limit as x approaches 0 of (1+x)^2-1 ?
The limit as x approaches 0 of (1+x)^2-1 is 0