lim tan x-sinx/x3 x--0

Given , limx→0 tanx-sinxx3After putting limit we are getting 00 form so using L'hospital rule , limx→0 ddxtanx-sinxddxx3=limx→0 sec2x-cos x3x2Again using L'hospital rule we get, limx→0 ddxsec2x-cos xddx3x2=limx→0 2secx secxtanx+sinx6x=limx→0 2sinxcos3x+sinx6x=limx→02cos3xsinxx+sinxx6Now as we know that limx→0sinxx=1 so putting the limit we get, =2+16=12

lim tan x-sinx/x³

x--0

Given, \lim_{x \to 0} \frac{tanx - sinx}{x^{3}} After putting limit we are getting \frac{0}{0} form so using L' hospital rule, \lim_{x \to 0} \frac{\frac{d}{dx} (tanx - sinx)}{\frac{d}{dx} x^{3}} = \lim_{x \to 0} \frac{\sec^{2} x - \cos x}{3 x^{2}} Again using L' hospital rule we get, \lim_{x \to 0} \frac{\frac{d}{dx} (\sec^{2} x - \cos x)}{\frac{d}{dx} 3 x^{2}} = \lim_{x \to 0} \frac{2 secx . secxtanx + sinx}{6 x} = \lim_{x \to 0} \frac{2 \frac{sinx}{\cos^{3} x} + sinx}{6 x} = \lim_{x \to 0} \frac{\frac{2}{\cos^{3} x} \frac{sinx}{x} + \frac{sinx}{x}}{6} Now as we know that \lim_{x \to 0} \frac{sinx}{x} = 1 so putting the limit we get, = \frac{2 + 1}{6} = \frac{1}{2}

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