Question

lim x=0 (sqrt(1+x) -1 )/x

Answer 1

1/2

Answer 2

The given limit is of the indeterminate form $\frac{0}{0}$ when $x=0$. To evaluate this, we use the method of rationalization. We multiply the numerator and the denominator by the conjugate of the numerator, which is $(\sqrt{1+x} + 1)$. This eliminates the square root in the numerator, allowing us to simplify the expression and then substitute the limit value.

The given limit is: $\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}$ When we substitute $x=0$, we get $\frac{\sqrt{1+0} - 1}{0} = \frac{1-1}{0} = \frac{0}{0}$, which is an indeterminate form.

To resolve this, multiply the numerator and denominator by the conjugate of the numerator, $(\sqrt{1+x} + 1)$: $\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \times \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1}$ Using the difference of squares formula, $(a-b)(a+b) = a^2 - b^2$, for the numerator: $= \lim_{x \to 0} \frac{(\sqrt{1+x})^2 - 1^2}{x(\sqrt{1+x} + 1)}$ $= \lim_{x \to 0} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)}$ $= \lim_{x \to 0} \frac{x}{x(\sqrt{1+x} + 1)}$ Since $x \to 0$, $x \neq 0$, we can cancel out $x$ from the numerator and denominator: $= \lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1}$ Now, substitute $x=0$ into the simplified expression: $= \frac{1}{\sqrt{1+0} + 1}$ $= \frac{1}{\sqrt{1} + 1}$ $= \frac{1}{1 + 1}$ $= \frac{1}{2}$