Question

〖lim〗┬(x→0) (√(1+x)-1)/x

Answer 1

$\lim_{x \to 0} \frac{\sqrt{1+x} -1}{x} = \lim_{x \to 0} \frac{\sqrt{1+x} -1}{(1+x)-1} = \\\\ =\lim_{x \to 0} \frac{\sqrt{1+x} -1}{(\sqrt{1+x} -1)(\sqrt{1+x} +1)} =\lim_{x \to 0} \frac{1}{\sqrt{1+x} +1}=\frac{1}{2}$

Answer 2

To find the limit of the given expression as x approaches 0, we can use L'Hôpital's rule. L'Hôpital's rule allows us to find the limit of an indeterminate form of the type 0/0 or ∞/∞ by taking the derivative of the numerator and the derivative of the denominator and then finding the limit of the resulting expression.

Let's apply L'Hôpital's rule to the given expression:

lim┬(x→0) (√(1+x) - 1) / x

First, find the derivative of the numerator and denominator separately:

Derivative of √(1+x) - 1 with respect to x: d/dx (√(1+x) - 1) = 1 / (2√(1+x))

Derivative of x with respect to x: d/dx (x) = 1

Now, take the limit of the new expression as x approaches 0:

lim┬(x→0) [1 / (2√(1+x))] / 1

At x = 0, the denominator becomes 2√(1+0) = 2√1 = 2, and the numerator becomes 1 / 2√(1+0) = 1 / 2√1 = 1 / 2.

So, the limit is:

lim┬(x→0) (√(1+x) - 1) / x = 1 / 2.

Therefore, the limit of the given expression as x approaches 0 is 1/2.