11. rationalising
we have
√1-sin2x(Ï€-4x)*1/√1+√sin2x
=|cosx-sinx| / (Î -4x)(2)
as x<Î /4
therefore it equals
=(cosx-sinx)/2(Î -4x)
as zero by zero form apply l hospitals rule
=-(sinx+cosx)/2(-4)
=root2 /8
=1/4√2
1. \lim_{x\rightarrow \frac{\prod{}}{4}} \frac{\sqrt{1-\sqrt{sin 2x}}}{\prod{-4x}} where x <pi/4
2. \lim_{x\rightarrow \propto } x^{3} \left\{{\sqrt{x^{2}+\sqrt{1+x^{4}}}- x\sqrt{2}} \right\}
3. \lim_{x\rightarrow \Pi /2} \frac{cos x}{(1-sinx)^{2/3}}
4. \lim_{x\rightarrow 0} \frac{\tan^{-1} x - \sin^{-1}x}{x^{3}}
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6 Answers
112.
rationalising
lim x→∞
x3 {x2 +√1+x4 -2x2 } /{√x2 +x2 √1+1x2 +x√2 }
lim x→∞
x3 {√1+x4-x2 }/2√2x
=
lim x→∞
x2 {1+x4-x4 }/2√2 (√1+x4 +x2}
=x2 /2√2x2 (√1+1x4 +1)
=
1/2√2(2)
=1/4√2
14. given is of from..........00
so
\lim_{x\rightarrow 0}\frac{\frac{1}{1+x^{2}}-\frac{1}{\sqrt{1-x^{2}}}}{3x^{2}}
= \lim_{x\rightarrow 0}\frac{\sqrt{1-x^{2}}-(1+x^{2})}{3x^{2}(1+x^{2})\sqrt{1-x^{2}}}
= \lim_{x\rightarrow 0}\frac{{1-x^{2}}-(1+x^{2})^{2}}{3x^{2}(1+x^{2})\sqrt{1-x^{2}}\left[ \sqrt{1-x^{2}}+(1+x^{2})\right]}
= \lim_{x\rightarrow 0}\frac{-3-x^{2}}{3(1+x^{2})\sqrt{1-x^{2}}\left[ \sqrt{1-x^{2}}+(1+x^{2})\right]}
= -36 = -12
