91Ó°ÊÓ

#tag_title#Step 3: Use the double-angle identity for sine#tag_content# We notice that the denominator contains a sine double-angle term. Let's use the double-angle identity for sine, which states that \(sin(2x) = 2sin(x)cos(x)\). Now, the expression becomes: $$ \lim _{x \rightarrow 0} \frac{6x^{2}\cos x}{\sin^{2}x \cdot 2\sin x\cos x} \cdot \cos(\cos x+\pi \tan (\frac{\pi}{4 \sec x})-1) $$ Simplify the expression by canceling some terms: $$ \lim _{x \rightarrow 0} \frac{3x^{2}}{\sin^{3}x} \cdot \cos(\cos x+\pi \tan (\frac{\pi}{4 \sec x})-1) $$ #tag_title#Step 4: Apply L'Hopital's Rule to the first part of the expression#tag_content# Now, let's consider the first part of the expression, \(\lim_{x \rightarrow 0} \frac{3x^{2}}{\sin^{3}x}\). As x approaches 0, the numerator approaches 0 and the denominator approaches 0 as well, which is in the indeterminate form \(\frac{0}{0}\). In this case, we can apply L'Hopital's Rule. We need to take the derivative of the numerator and the denominator with respect to x: $$ \frac{d}{dx}(3x^{2}) = 6x $$ and $$ \frac{d}{dx}(\sin^{3}x) = 3\sin^{2}x \cdot \cos x $$ Applying L'Hopital's Rule, we get: $$ \lim _{x \rightarrow 0} \frac{6x}{3\sin^{2}x \cdot \cos x} $$ Simplify the expression: $$ \lim _{x \rightarrow 0} \frac{2x}{\sin^{2}x \cdot \cos x} $$ #tag_title#Step 5: Evaluate the limits separately#tag_content# Now, we can split the given expression into two separate limits to evaluate their limit separately: $$ \left(\lim _{x \rightarrow 0} \frac{2x}{\sin^{2}x \cdot \cos x}\right) \cdot \left(\lim_{x \rightarrow 0} \cos(\cos x+\pi \tan (\frac{\pi}{4 \sec x})-1)\right) $$ We've already determined the first limit in step 4. Now let's evaluate the second limit. As x approaches 0, the inner argument of cosine approaches \(cos(0) + \pi\tan(\frac{\pi}{4}) - 1\), which simplifies to \(1 - 1 = 0\). And since \(\cos(0) = 1\), the second limit is 1: $$ \left(\lim _{x \rightarrow 0} \frac{2x}{\sin^{2}x \cdot \cos x}\right) \cdot 1 $$ #tag_title#Step 6: Evaluate the final limit#tag_content# Now we need to evaluate the remaining limit: $$ \lim _{x \rightarrow 0} \frac{2x}{\sin^{2}x \cdot \cos x} $$ Let's apply L'Hopital's Rule again to this expression. We take the derivatives of the numerator and denominator with respect to x: $$ \frac{d}{dx}(2x) = 2 $$ and $$ \frac{d}{dx}(\sin^{2}x \cdot \cos x) = 4\sin x \cdot \cos^{2}x - \sin^{3} x $$ Now, we get the following limit after applying L'Hopital's Rule: $$ \lim _{x \rightarrow 0} \frac{2}{4\sin x \cdot \cos^{2}x - \sin^{3} x} $$ As x approaches 0, the denominator approaches \(4(0)(1) - 0 = 0\). So, the limit is in the indeterminate form \(\frac{2}{0}\). However, this particular expression doesn't have a limit as x approaches 0. The reason is that the denominator oscillates between positive and negative values with increasing speed as x gets closer to 0, which makes the whole expression undefined at the limit point. Therefore, the answer is: $$ \lim _{x \rightarrow 0} \frac{2x}{\sin^{2}x \cdot \cos x} \cdot \cos(\cos x+\pi \tan (\frac{\pi}{4 \sec x})-1) = \text{Undefined} $$