Question

Evaluate the limit: $\lim_{x \to 1} \left[ \frac{\sqrt{x} - 1}{\log x} \right] =$

• A. $\frac{1}{2}$
• B. 2
• C. $-2$
• D. $-\frac{1}{2}$

Hint:

You can also solve this using algebraic substitution! Let $x = t^2$. As $x \to 1$, $t \to 1$. The limit transforms into $\lim_{t \to 1} \frac{t-1}{\log(t^2)} = \lim_{t \to 1} \frac{t-1}{2\log t}$. Since $\lim_{t \to 1} \frac{\log t}{t-1} = 1$ is a standard fundamental limit property, its reciprocal is also 1, leaving you with just the constant factor $\frac{1}{2}$ in front.

Answer 1

The Correct Option is A

Step 1: Test direct substitution.
Putting $x = 1$ gives $\dfrac{\sqrt{1} - 1}{\log 1} = \dfrac{0}{0}$, an indeterminate form, so we reshape rather than substitute.
Step 2: Shift to a small variable.
Let $x = 1 + t$ with $t \to 0$. The limit becomes $\displaystyle\lim_{t \to 0}\dfrac{\sqrt{1+t} - 1}{\log(1+t)}$.
Step 3: Approximate the numerator.
For small $t$, $\sqrt{1+t} = (1+t)^{1/2} \approx 1 + \dfrac{t}{2}$, so $\sqrt{1+t} - 1 \approx \dfrac{t}{2}$.
Step 4: Approximate the denominator.
The standard expansion gives $\log(1+t) \approx t$ for small $t$.
Step 5: Form the ratio.
$\dfrac{\frac{t}{2}}{t} = \dfrac{1}{2}$, independent of $t$.
Step 6: Take the limit.
As $t \to 0$ the value stays $\dfrac{1}{2}$.
\[ \boxed{\dfrac{1}{2}} \]