Direct answer

The limit \(\lim_{x\to 0}\frac{2-\sqrt{4-x}}{x}\) exists and its value is \(-\dfrac{1}{4}\).

Introduction

The limit \(\lim_{x\to 0}\frac{2-\sqrt{4-x}}{x}\) is a classic calculus problem that tests understanding of algebraic manipulation and the
connection between limits and derivatives.

This article walks through a detailed solution using rationalization and then shows an elegant interpretation via the derivative of \(\sqrt{4-x}\).

Step‑by‑step solution of the limit

1. Write the problem

Consider the limit

\[
L=\lim_{x\to 0}\frac{2-\sqrt{4-x}}{x}.
\]

Direct substitution \(x=0\) gives \(\frac{2-\sqrt{4}}{0}=\frac{0}{0}\), which is an indeterminate form that requires simplification.

2. Rationalize the numerator

Multiply numerator and denominator by the conjugate \(2+\sqrt{4-x}\):

\[
\frac{2-\sqrt{4-x}}{x}\cdot\frac{2+\sqrt{4-x}}{2+\sqrt{4-x}}
=\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{x(2+\sqrt{4-x})}.
\]

Using \((a-b)(a+b)=a^{2}-b^{2}\):

\[
(2-\sqrt{4-x})(2+\sqrt{4-x})=2^{2}-(\sqrt{4-x})^{2}=4-(4-x)=x.
\]

So, for \(x \ne 0\),

\[
\frac{2-\sqrt{4-x}}{x}
=\frac{x}{x(2+\sqrt{4-x})}
=\frac{1}{2+\sqrt{4-x}}.
\]

3. Carefully track the sign

Note that the original expression can also be written as
\[
\frac{2-\sqrt{4-x}}{x}=-\frac{\sqrt{4-x}-2}{x}.
\]

Rationalizing \(\sqrt{4-x}-2\) gives
\[
-\frac{\sqrt{4-x}-2}{x}
=-\frac{-(x)}{x(\sqrt{4-x}+2)}
=-\frac{1}{\sqrt{4-x}+2}.
\]

4. Evaluate the limit

Now compute

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

Therefore, the value of the limit is

\[
\boxed{-\dfrac{1}{4}}.
\]

Interpretation using derivatives

The given limit can be viewed through the definition of the derivative of the function \(f(x)=\sqrt{4-x}\) at \(x=0\).

The derivative by first principles is
\[
f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}
=\lim_{h\to 0}\frac{\sqrt{4-h}-2}{h}.
\]

Observe that
\[
\frac{2-\sqrt{4-x}}{x}=-\frac{\sqrt{4-x}-2}{x},
\]
so the original limit equals \(-f'(0)\).

From the standard derivative of \(\sqrt{4-x}\),
\[
f'(x)=-\frac{1}{2\sqrt{4-x}},
\]
which gives \(f'(0)=-\frac{1}{4}\) and hence \(-f'(0)=-\frac{1}{4}\), confirming the previous result.

Key points for exam practice

• Always check for the indeterminate form \(0/0\) when evaluating limits involving square roots.
• Use conjugates to simplify expressions like \(\sqrt{a-x}-b\) or \(b-\sqrt{a-x}\) before taking the limit.
• Recognize when a limit matches the form of a derivative, as this often provides a faster method and deeper conceptual understanding.