JEE Main 2020 · 3 September · Morning Shift · Mathematics · Limits
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limx→0{1x8(1−cosx22−cosx24+cosx22cosx24)}\lim_{x \to 0}\left\{\dfrac{1}{{x}^{8}}\left(1 - cos\dfrac{{x}^{2}}{2} - cos\dfrac{{x}^{2}}{4} + cos\dfrac{{x}^{2}}{2}cos\dfrac{{x}^{2}}{4}\right)\right\}limx→0{x81(1−cos2x2−cos4x2+cos2x2cos4x2)}
=limx→0(1−cosx22)(1−cosx24)x8= \lim_{x \to 0}\dfrac{\left(1 - cos\dfrac{{x}^{2}}{2}\right)\left(1 - cos\dfrac{{x}^{2}}{4}\right)}{{x}^{8}}=limx→0x8(1−cos2x2)(1−cos4x2)
=limx→0(1−cosx22)x4×(1−cosx24)x4= \lim_{x \to 0}\dfrac{\left(1 - cos\dfrac{{x}^{2}}{2}\right)}{{x}^{4}} \times \dfrac{\left(1 - cos\dfrac{{x}^{2}}{4}\right)}{{x}^{4}}=limx→0x4(1−cos2x2)×x4(1−cos4x2)
=limx→0(1−cosx22)4×(x44)×(1−cosx24)16×(x416)= \lim_{x \to 0}\dfrac{\left(1 - cos\dfrac{{x}^{2}}{2}\right)}{4 \times \left(\dfrac{{x}^{4}}{4}\right)} \times \dfrac{\left(1 - cos\dfrac{{x}^{2}}{4}\right)}{16 \times \left(\dfrac{{x}^{4}}{16}\right)}=limx→04×(4x4)(1−cos2x2)×16×(16x4)(1−cos4x2)
Using limx→01−cosxx2=12\lim_{x \to 0}\dfrac{1 - \cos x}{{x}^{2}} = \dfrac{1}{2}limx→0x21−cosx=21
=12×14×12×116=1256=2−8= \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{1}{2} \times \dfrac{1}{16} = \dfrac{1}{256} = {2}^{- 8}=21×41×21×161=2561=2−8
⇒2−8=2−k⇒k=8⇒ {2}^{- 8} = {2}^{- k} ⇒ k = 8⇒2−8=2−k⇒k=8
