Question

$\lim_{x \to 0} \frac{\sin(6x^2)}{ln \cos(2x^2 - x)}$ is equal to

• A. 12
• B. -12
• C. 6
• D. -6

Answer 1

Answer: (B)

-12

Answer 2

To evaluate the limit $\lim_{x \to 0} \frac{\sin(6x^2)}{\ln \cos(2x^2 - x)}$, we first check the form of the limit by substituting $x=0$:

Numerator: $\sin(6(0)^2) = \sin(0) = 0$

Denominator: $\ln \cos(2(0)^2 - 0) = \ln \cos(0) = \ln(1) = 0$

The limit is of the indeterminate form $\frac{0}{0}$. We can use standard limit formulas.

We use the following standard limits and approximations for small $y$:

1. $\lim_{y \to 0} \frac{\sin y}{y} = 1 \implies \sin y \approx y$

2. $\lim_{y \to 0} \frac{\ln(1+y)}{y} = 1 \implies \ln(1+y) \approx y$

3. $\lim_{y \to 0} \frac{1-\cos y}{y^2} = \frac{1}{2} \implies 1-\cos y \approx \frac{y^2}{2} \implies \cos y - 1 \approx -\frac{y^2}{2}$

Let's apply these to the given expression:

Numerator:

$\sin(6x^2)$

As $x \to 0$, $6x^2 \to 0$. Using approximation $\sin y \approx y$:

$\sin(6x^2) \approx 6x^2$

Denominator:

$\ln \cos(2x^2 - x)$

We rewrite $\cos(2x^2 - x)$ as $1 + (\cos(2x^2 - x) - 1)$.

Let $Y = \cos(2x^2 - x) - 1$. As $x \to 0$, $2x^2 - x \to 0$, so $\cos(2x^2 - x) \to \cos(0) = 1$, which means $Y \to 0$.

Using approximation $\ln(1+Y) \approx Y$:

$\ln \cos(2x^2 - x) = \ln(1 + (\cos(2x^2 - x) - 1)) \approx \cos(2x^2 - x) - 1$

Now, we need to approximate $\cos(2x^2 - x) - 1$.

Let $\theta = 2x^2 - x$. As $x \to 0$, $\theta \to 0$.

Using approximation $\cos \theta - 1 \approx -\frac{\theta^2}{2}$:

$\cos(2x^2 - x) - 1 \approx -\frac{(2x^2 - x)^2}{2}$

Substitute these approximations back into the limit expression:

$\lim_{x \to 0} \frac{\sin(6x^2)}{\ln \cos(2x^2 - x)} = \lim_{x \to 0} \frac{6x^2}{-\frac{(2x^2 - x)^2}{2}}$

Simplify the denominator:

$(2x^2 - x)^2 = (x(2x - 1))^2 = x^2(2x - 1)^2$

So the limit becomes:

$\lim_{x \to 0} \frac{6x^2}{-\frac{x^2(2x - 1)^2}{2}}$

Cancel out $x^2$ from the numerator and denominator (since $x \ne 0$ as $x \to 0$):

$= \lim_{x \to 0} \frac{6}{-\frac{(2x - 1)^2}{2}}$

Now, substitute $x=0$:

$= \frac{6}{-\frac{(2(0) - 1)^2}{2}}$

$= \frac{6}{-\frac{(-1)^2}{2}}$

$= \frac{6}{-\frac{1}{2}}$

$= 6 \times (-2)$

$= -12$

The final answer is $\boxed{-12}$.