X=0. (_log(1-x+x+)-log(1-x=x?) Let ...
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X=0. (_log(1-x+x+)-log(1-x=x?) Let Secx-cosx Then the value of f(0) so that fis continuous at x=0 is:

JEE/Engineering Exams
Maths
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( f(x)=frac{log left(1+x+x^{2}right)+log left(1-x+x^{2}right)}{sec x-cos x} ) ( f(x)=operatorname{lif}_{x rightarrow 0} f(x)=operatorname{st.} frac{log left(1+x+x^{2}right)+operatorname{cog}left(1-x+x^{2}right)}{sec x-cos x} ) ( =sec _{x rightarrow 0} frac{log left(1+x+x^{2}right)+log left(1-x+x^{2}right)}{tan x sin x} ) Divide by ( x^{2} ) ( =frac{Delta t}{x-10} cdot frac{log left(1+x+x^{2}right)}{frac{x^{2}}{tan x} cdot frac{operatorname{cog}left(1-x+x^{2}right)}{x^{2}}} ) ( thereforeleft[f(0)=frac{1}{1}right. )
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