Question

lim x стремится к 0 ((корень из 1+x)-1)\sin(пи(x+2))... получается неопределенность вида 0\0 помогите пожалуйста, очень плохо данную неопределенность раскрываю

Answer 1

$lim_{x->0} \frac{\sqrt{1+x}-1}{sin(\pi(x+2))} =\\\\ lim_{x->0} \frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{sin(\pi*x+2*\pi)*(\sqrt{1+x}+1)} =\\\\ lim_{x->0} \frac{1+x-1}{sin(\pi*x)*(\sqrt{1+x}+1)} =\\\\ lim_{x->0} \frac{x}{sin(\pi*x)*(\sqrt{1+x}+1)} =\\\\ lim_{x->0} \frac{\pi*x}{sin(\pi*x)*\pi*(\sqrt{1+x}+1)} =\\\\ \frac{1}{\pi*\sqrt{1+0}+1}=\frac{1}{2*\pi}$

Answer 2

все очень просто, все решение есть

Answer 3

Solving the Indeterminate Form

To solve the indeterminate form of the given expression, we can use L'Hôpital's rule, which states that for functions f(x) and g(x) differentiable on an open interval containing a, except possibly at a itself, if:

1. lim(x→a) f(x) = 0 and lim(x→a) g(x) = 0, or 2. lim(x→a) f(x) = ±∞ and lim(x→a) g(x) = ±∞,

then:

lim(x→a) (f(x) / g(x)) = lim(x→a) (f'(x) / g'(x)), provided the limit on the right-hand side exists or is ±∞.

Applying L'Hôpital's Rule

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

lim(x→0) ((√(1+x) - 1) / sin(π(x+2)))

First, we need to check if the conditions for L'Hôpital's rule are met. We have lim(x→0) (√(1+x) - 1) = 0 and lim(x→0) sin(π(x+2)) = 0, so the conditions are satisfied.

Now, we can differentiate the numerator and the denominator separately and then take the limit again.

Differentiating the Numerator and Denominator

The derivative of the numerator is:

((1/2)*(1+x)^(-1/2))

The derivative of the denominator is:

π*cos(π(x+2))

Taking the Limit Again

Now, we can take the limit of the ratio of the derivatives as x approaches 0:

lim(x→0) (((1/2)*(1+x)^(-1/2)) / (π*cos(π(x+2))))

Final Result

After applying L'Hôpital's rule, the final result of the limit can be obtained by evaluating the limit of the ratio of the derivatives. This will provide the value of the original limit.

I hope this helps! If you have any further questions or need additional assistance, feel free to ask.