Question

Lim sin^2 4x/6x^2. X->0

Answer 1

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Answer 2

To find the limit of the function $f(x) = \frac{{\sin^2(4x)}}{{6x^2}}$ as $x$ approaches 0, we can apply some trigonometric identities and algebraic simplifications.

Let's start by using the identity $\sin^2(x) = \frac{{1 - \cos(2x)}}{2}$. Substituting this into our function, we get:

$f(x) = \frac{{\frac{{1 - \cos(8x)}}{2}}}{{6x^2}}$

Now, we can simplify further by canceling out the common factors of 2 in the numerator and denominator:

$f(x) = \frac{{1 - \cos(8x)}}{{12x^2}}$

Next, let's deal with the $\cos(8x)$ term. We know that $\cos(0) = 1$ and $\lim_{{x \to 0}} \frac{{\sin(x)}}{x} = 1$ (a common limit in trigonometry). We can use these facts to rewrite $\cos(8x)$ as follows:

$\cos(8x) = 1 - 2\sin^2(4x)$

Substituting this back into our function:

$f(x) = \frac{{1 - (1 - 2\sin^2(4x))}}{{12x^2}}$

Simplifying further:

$f(x) = \frac{{2\sin^2(4x)}}{{12x^2}}$

Now, we can cancel out the factor of 2:

$f(x) = \frac{{\sin^2(4x)}}{{6x^2}}$

Finally, we can use the trigonometric identity $\lim_{{x \to 0}} \frac{{\sin(x)}}{x} = 1$ to evaluate the limit:

$\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} \frac{{\sin^2(4x)}}{{6x^2}} = \frac{1}{6}$

Therefore, the limit of the given function as $x$ approaches 0 is $\frac{1}{6}$.