〖lim〗┬(x→0) (√(1+x)-1)/x

Предыдущий
вопрос Следующий
вопрос

To find the limit as x approaches 0 of (√(1+x) - 1) / x, we can use L'Hopital's Rule.

Let f(x) = √(1+x) - 1 and g(x) = x.
Therefore, f(0) = 0 and g(0) = 0.

Taking the derivative of f(x) and g(x) gives us:

f'(x) = 1 / (2√(1+x))
g'(x) = 1

Now, we can apply L'Hopital's Rule:

lim┬(x→0) √(1+x) - 1 / x = lim┬(x→0) f(x) / g(x)
= lim┬(x→0) f'(x) / g'(x)
= lim┬(x→0) 1 / (2√(1+x))

Now, plug in x = 0:

lim┬(x→0) 1 / (2√(1+0))
= 1 / (2)

Therefore, the limit of (√(1+x) - 1) / x as x approaches 0 is 1/2.