Question

\[ \lim_{x\to 0} \frac{\sqrt{1+x}-1}{x} \] is ____________ (rounded off to one decimal place).

Hint:

When a limit gives a $0/0$ form, try rationalizing the numerator or denominator.

Answer 1

Correct Answer: 3

Rationalize the numerator:
\[ \frac{\sqrt{1+x}-1}{x} = \frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)}. \]
Simplify using \[ (\sqrt{1+x}-1)(\sqrt{1+x}+1) = (1+x)-1 = x. \]
So the expression becomes:
\[ \frac{x}{x(\sqrt{1+x}+1)} = \frac{1}{\sqrt{1+x}+1}. \]
Now take the limit as \(x \to 0\):
\[ \lim_{x\to 0} \frac{1}{\sqrt{1+x}+1} = \frac{1}{1+1} = 0.5. \]
Thus,
\[ \boxed{0.5} \]