### Related Formula To evaluate limits of indeterminate types containing radical forms, rationalize the numerator directly by multiplying by its conjugate element matching: (sqrtA - sqrtB)(sqrtA + sqrtB) = A - B$(\sqrt{A} - \sqrt{B})(\sqrt{A} + \sqrt{B}) = A - B$ ### Core Logic Rewrite the expression as a fraction with sin x$\sin x$ in the denominator and rationalize the numerator: lim_x to 0 frac(2cos^2 x + 3cos x) - (cos^2 x + sin x + 4)sin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$\lim_{x \to 0} \frac{(2\cos^2 x + 3\cos x) - (\cos^2 x + \sin x + 4)}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$ = lim_x to 0 fraccos^2 x + 3cos x - sin x - 4sin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$= \lim_{x \to 0} \frac{\cos^2 x + 3\cos x - \sin x - 4}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$ ### Step 1: Simplify Numerator and Group Terms Express the numerator terms to isolate algebraic patterns: cos^2 x + 3cos x - 4 - sin x = (cos x - 1)(cos x + 4) - sin x$\cos^2 x + 3\cos x - 4 - \sin x = (\cos x - 1)(\cos x + 4) - \sin x$ Substitute this back into our rationalized limit format: = lim_x to 0 frac(cos x - 1)(cos x + 4) - sin xsin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$= \lim_{x \to 0} \frac{(\cos x - 1)(\cos x + 4) - \sin x}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$ ### Step 2: Distribute sin x$\sin x$ in Denominator Split the limit across the two separated numerator expressions: = lim_x to 0 left[ fraccos x - 1sin x cdot (cos x + 4) - 1 right] cdot frac1sqrt2(1)+3 + sqrt1+0+4$= \lim_{x \to 0} \left[ \frac{\cos x - 1}{\sin x} \cdot (\cos x + 4) - 1 \right] \cdot \frac{1}{\sqrt{2(1)+3} + \sqrt{1+0+4}}$ Evaluate the limit component values: lim_x to 0 fraccos x - 1sin x = lim_x to 0 frac-2sin^2(x/2)2sin(x/2)cos(x/2) = lim_x to 0 [-tan(x/2)] = 0$\lim_{x \to 0} \frac{\cos x - 1}{\sin x} = \lim_{x \to 0} \frac{-2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)} = \lim_{x \to 0} [-\tan(x/2)] = 0$ Substituting this zero value simplifies the numerator expression directly: = left[ 0 cdot (1 + 4) - 1 right] cdot frac1sqrt5 + sqrt5 = frac-12sqrt5$= \left[ 0 \cdot (1 + 4) - 1 \right] \cdot \frac{1}{\sqrt{5} + \sqrt{5}} = \frac{-1}{2\sqrt{5}}$ ### Pattern Recognition Recognizing that fraccos x - 1sin x to 0$\frac{\cos x - 1}{\sin x} \to 0$ as x to 0$x \to 0$ isolates the non-vanishing trigonometric components without needing full multi-stage application of L'Hôpital's rule. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limits and Derivatives