Question

$\lim_{x \to 1} \frac{\log x}{x-1}=$

• A. 1
• B. -1
• C. 0
• D. $\infty$

Answer 1

Answer: (A)

1

Answer 2

The given limit is $\lim_{x \to 1} \frac{\log x}{x-1}$.

First, let's check the form of the limit by substituting $x=1$: Numerator: $\log x \to \log 1 = 0$ Denominator: $x-1 \to 1-1 = 0$ The limit is of the indeterminate form $\frac{0}{0}$.

We can evaluate this limit using L'Hôpital's Rule or by using a standard limit formula.

Method 1: Using L'Hôpital's Rule

Since the limit is of the $\frac{0}{0}$ form, we can apply L'Hôpital's Rule, which states that if $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Let $f(x) = \log x$ and $g(x) = x-1$. Then, find their derivatives: $f'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}$ $g'(x) = \frac{d}{dx}(x-1) = 1$

Now, apply L'Hôpital's Rule:

$$\lim_{x \to 1} \frac{\log x}{x-1} = \lim_{x \to 1} \frac{\frac{1}{x}}{1}$$ $$= \lim_{x \to 1} \frac{1}{x}$$

Substitute $x=1$ into the expression:

$$= \frac{1}{1} = 1$$

Method 2: Using Standard Limit Formula

Let $x-1 = h$. As $x \to 1$, $h \to 0$. From $x-1=h$, we get $x = 1+h$. Substitute these into the limit expression:

$$\lim_{x \to 1} \frac{\log x}{x-1} = \lim_{h \to 0} \frac{\log(1+h)}{h}$$

This is a standard limit formula, which is known to be equal to 1:

$$\lim_{h \to 0} \frac{\log(1+h)}{h} = 1$$

Both methods yield the same result.