Tìm lim

To find the limit

\[
\lim_{x \to 0} \frac{\sqrt{1 + 2x} - x - 4}{x^2}
\]

we can start by substituting \(x = 0\):

\[
\sqrt{1 + 2(0)} - 0 - 4 = \sqrt{1} - 0 - 4 = 1 - 4 = -3
\]

So the expression becomes:

\[
\frac{-3}{0}
\]

Since this leads to an indeterminate form, we will need to apply L'Hôpital's Rule. This rule states that if we encounter a limit that gives a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form, we can take the derivative of the numerator and the derivative of the denominator separately.

1. **Differentiate the numerator**:
- The derivative of \( \sqrt{1 + 2x} \) is \( \frac{1}{2\sqrt{1 + 2x}} \cdot 2 = \frac{1}{\sqrt{1 + 2x}} \).
- The derivative of \( -x \) is \(-1\).
- The derivative of \(-4\) is \(0\).

Thus, the derivative of the numerator is:

\[
\frac{1}{\sqrt{1 + 2x}} - 1
\]

2. **Differentiate the denominator** \( x^2 \):
The derivative is \(2x\).

Now we apply L'Hôpital's Rule:

\[
\lim_{x \to 0} \frac{\frac{1}{\sqrt{1 + 2x}} - 1}{2x}
\]

Substituting \(x = 0\) into the new limit gives:

\[
\frac{\frac{1}{\sqrt{1 + 0}} - 1}{2(0)} = \frac{1 - 1}{0} = \frac{0}{0}
\]

Since we still have an indeterminate form, we apply L'Hôpital's Rule again.

1. **Differentiate the numerator again**:
\[
\frac{d}{dx}\left( \frac{1}{\sqrt{1 + 2x}} - 1 \right) = -\frac{1}{2(1 + 2x)^{3/2}} \cdot 2 = -\frac{1}{(1 + 2x)^{3/2}}
\]

2. **Differentiate the denominator**:
The derivative of \(2x\) is \(2\).

Now, we evaluate the limit:

\[
\lim_{x \to 0} \frac{-\frac{1}{(1 + 2x)^{3/2}}}{2}
\]

Substituting \(x = 0\):

\[
-\frac{1}{(1 + 0)^{3/2} \cdot 2} = -\frac{1}{1 \cdot 2} = -\frac{1}{2}
\]

Therefore, the final result is:

\[
\lim_{x \to 0} \frac{\sqrt{1 + 2x} - x - 4}{x^2} = -\frac{1}{2}
\]