Dear student limx→01+tan2x3xlimx→0e3xln1+tan2xIf limu→bf(u)=L and limx→ag(x)=b and f(x) is continuous at x=b then limx→af(g(x))=LSo, g(x)=3xln1+tan2xApply L'Hoiptal's rule, we get3limx→0sec2x tanxxtan2x+11=3limx→0tanxxcos2xtan2x+1Apply L'Hopital's rule, we get3limx→0sec2x2xcos2xtan2x+12x+-sin2xtan2x+1+2sec2xcos2xtanx2=3limx→01cos2xtan2x cos2x +cos2x-xsin2xtan2x-xsin2x+2xtanx=3So, limu→3+eu=e3

How to solve Q43

How to solve Q43 39. Three points are chosen randomly and independently on a circle. What is the robability that all three pairwise distances tween the points are less than the radius of tan-i 45. lim is equal to circle? (C) 1/18 (b) 1t24 (d) 1/12 40. A line segment with the end points A (3, —2) and B (6, 4) is divided into three equal parts. Find the coordinates of the division points. Jay-(4. 0). (5.2) (d) (0.4), (2, 5) Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of the larger sphere? 1 — COS4x 46. lim is equal to x-•0 2sin2 x + x tan 7x (d) '17. If then lim uals (a) -715 -7 14 30 2 123 4 3 9 2 (1 —cos 42. lim is equal to x-.R/4tan2x—sin x 43. lim (1 + tan2 G)3/x is equal to 3 log ax — 1 44. lim is equal to 48. 49. Given the vertices of a triangle are A (I, —1, —3) B (2, 1, —2) and C (—5, 2, —6). Compute th' length of the bisector of the interior angle vertex A. 3Jiö (d) 3Viö 4 It is known that AB=2a —6b and AC=3a + where a and b are mutually perpendicul unit vectors. Determine the angles of th AABC. 50. Find the component of the vector a (—1,2, perpendicular to the plane Of the vector et (1, O, 1) and (1, 1, 1). (c) loga e (b) Adflloga (a) (1/2, O, 1/2) (c) (1/2, o, -1/2) (b) (-1/2, O, 1/2) (d) (-1/2, O, -1/2)

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\lim_{x \to 0} {(1 + \tan^{2} \sqrt{x})}^{\frac{3}{x}} \lim_{x \to 0} e^{\frac{3}{x} \ln (1 + \tan^{2} \sqrt{x})} If \lim_{u \to b} f (u) = L and \lim_{x \to a} g (x) = b and f (x) is continuous at x = b . then \lim_{x \to a} f (g (x)) = L So, g (x) = \frac{3}{x} \ln (1 + \tan^{2} \sqrt{x}) Apply L' Hoiptal' s rule, we get 3 \lim_{x \to 0} (\frac{\frac{\sec^{2} \sqrt{x} \tan \sqrt{x}}{\sqrt{x} (\tan^{2} \sqrt{x} + 1)}}{1}) = 3 \lim_{x \to 0} (\frac{\tan \sqrt{x}}{\sqrt{x} \cos^{2} \sqrt{x} (\tan^{2} \sqrt{x} + 1)}) Apply L' Hopital' s rule, we get 3 \lim_{x \to 0} (\frac{\frac{\sec^{2} \sqrt{x}}{2 \sqrt{x}}}{\frac{\cos^{2} \sqrt{x} (\tan^{2} \sqrt{x} + 1)}{2 \sqrt{x}} + \frac{- \sin (2 \sqrt{x}) (\tan^{2} \sqrt{x} + 1) + 2 \sec^{2} \sqrt{x} \cos^{2} \sqrt{x} \tan \sqrt{x}}{2}}) = 3 \lim_{x \to 0} \frac{1}{\cos^{2} (\sqrt{x}) (\tan^{2} \sqrt{x} \cos^{2} \sqrt{x} + \cos^{2} \sqrt{x} - \sqrt{x} \sin (2 \sqrt{x}) \tan^{2} \sqrt{x} - \sqrt{x} \sin (2 \sqrt{x}) + 2 \sqrt{x} \tan \sqrt{x})} = 3 So, \lim_{u \to 3^{+}} e^{u} = e^{3}

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