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Lim\xa0x tends to 0\xa0(Tanx - sinx )/x3 |
| Answer» Ans.\xa0\\(\\lim_{x \\to 0} {tan x - sin x \\over x^3}\\)=>\xa0\\(\\lim_{x \\to 0} {{sin x\\over cos x} - sin x \\over x^3}\\)=>\xa0\\(\\lim_{x \\to 0} {sin x({1\\over cosx } - 1) \\over x^3}\\)=>\xa0\\(\\lim_{x \\to 0} {sin x(1 - cosx) \\over x^3 . \\space cos x }\\)=>\xa0\\(\\lim_{x \\to 0} {[{1\\over cos x}. {sin x \\over x} .{ 1 - cosx \\over x^2 }}]\\)=>\xa0\\(\\lim_{x \\to 0} {[{1\\over cos x}. {sin x \\over x} .{ 2sin^2{x\\over 2} \\over x^2 }}]\\)=>\xa0\\(2\\lim_{x \\to 0} {[{1\\over cos x}. {sin x \\over x} .{ sin{x\\over 2} \\over x}.{ sin{x\\over 2} \\over x}}]\\)=>\xa0\\(2\\lim_{x \\to 0} {[{1\\over cos x}. {sin x \\over x} .{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}.{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}}]\\)=>\xa0\\(2[\\lim_{x \\to 0} {{1\\over cos x}\\times \\lim_{x \\to 0}{sin x \\over x} \\times \\lim_{x \\to 0}{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}\\times \\lim_{x \\to 0}{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}}]\\)=>\xa0\\(2[\\lim_{x \\to 0} {{1\\over cos x}\\times \\lim_{x \\to 0}{sin x \\over x} \\times \\lim_{{x\\over 2} \\to 0}{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}\\times \\lim_{{x\\over2} \\to 0}{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}}]\\) [as x tends to zero, x/2 also tends to zero]=>\xa0\\(2\\times 1 \\times 1 \\times {1\\over 2}\\times {1\\over 2} = {1\\over 2}\\) | |