Question

Evaluate the limit \[ \lim_{x\to 0}\frac{\sqrt{\,1+x\,}-1}{x}. \]

• A. \(0\)
• B. \(\tfrac{1}{2}\)
• C. \(1\)
• D. \(2\)

Hint:

For radicals, three fast routes often agree: (i) rationalize, (ii) view as a derivative via the limit definition, or (iii) use a binomial/Taylor series near the expansion point. Pick whichever is quickest for the exam setting.

Answer 1

The Correct Option is B

Method 1 (Rationalization — algebraic).

Step 1: Identify the form. As \(x\to 0\), numerator \(\to 0\) and denominator \(\to 0\) \(\Rightarrow\) \(0/0\) indeterminate.

Step 2: Multiply by the conjugate. \[ \frac{\sqrt{1+x}-1}{x}\cdot\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1} =\frac{(1+x)-1}{x\big(\sqrt{1+x}+1\big)} =\frac{x}{x\big(\sqrt{1+x}+1\big)}. \]

Step 3: Cancel \(x\) (for \(x\neq 0\); the limit then follows by continuity). \[ =\frac{1}{\sqrt{1+x}+1}\xrightarrow[x\to 0]{}\frac{1}{1+1}=\frac{1}{2}. \] Method 2 (Derivative viewpoint — limit definition).
Let \(f(t)=\sqrt{t}\). Then \[ \frac{\sqrt{1+x}-1}{x} =\frac{f(1+x)-f(1)}{x} =\frac{f(1+x)-f(1)}{(1+x)-1} \xrightarrow[x\to 0]{} f'(1). \] Since \(f'(t)=\dfrac{1}{2\sqrt{t}}\), we get \(f'(1)=\dfrac{1}{2}\). Method 3 (Series expansion — binomial).
For \(|x|<1\), \[ \sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\cdots. \] Hence \[ \frac{\sqrt{1+x}-1}{x}=\frac{\frac{x}{2}-\frac{x^2}{8}+\cdots}{x} =\frac{1}{2}-\frac{x}{8}+\cdots \xrightarrow[x\to 0]{}\frac{1}{2}. \] Domain note. The expression is defined for \(x>-1\) (so that \(1+x\ge 0\)), which includes a deleted neighborhood of \(0\); therefore the limit process is valid.

Final Answer:
\[ \boxed{\tfrac{1}{2}} \]