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jishan
evaluate lim x->a ( 2- a/x)^tanpia/2x
take the log and use l'hospitals rule
in (1+ f(x))^(g(x)) , where f(x) ->0 and g(x) -> inf, mug it up that the final ans is e^( (f(x) g(x) )
solve step by step i m not understand.
Let \[\large y = ( 2-\frac{ a}{x})^{tan\frac{\pi a}{2x}}\]\[\ln\large y = \ln ( 2-\frac{ a}{x})^{tan\frac{\pi a}{2x}}\]\[\ln\large y = (tan\frac{\pi a}{2x})\ln( 2-\frac{ a}{x})\]Take limit on both sides\[\lim_{x\rightarrow a}(\ln\large y) = \lim_{x\rightarrow a}[(tan\frac{\pi a}{2x})\ln( 2-\frac{ a}{x})]\]Evaluate the limit on the right by l'hopital's rule. Then\[ \lim_{x\rightarrow a}( 2-\frac{ a}{x})^{tan\frac{\pi a}{2x}} = \lim_{x\rightarrow a} y = e^{\lim_{x\rightarrow a}lny}\], which is e^(the things you get after evaluating that limit)