We know that, #(1) : lim_(h to 0)(e^h-1)/h=lne=1, and #
#(2) : lim_(x to a)(x^n-a^n)/(x-a)=na^(n-1)#.
#"The Reqd. Lim.="lim_(x to 1)(e^x-e)/(x^50-1)#,
#=lim_(x to 1)(e(e^(x-1)-1))/(x-1)-:lim_(x to 1)(x^50-1)/(x-1)#,
#=l_1-:l_2.................(ast)," say, where, "#
#l_1=lim_(x to 1)(e(e^(x-1)-1))/(x-1)#.
Subst. #x-1=h," so that, as "x to 1, h to 0#.
#:. l_1=lim_(h to 0){e(e^h-1)}/h=e*1.............[because, (1)]#.
#rArr l_1=e.............................(ast^1)#.
Next, #l_2=lim_(x to 1)(x^50-1)/(x-1)#,
#=lim_(x to 1)(x^50-1^50)/(x-1)#,
#=50*1^(50-1)#.
#rArr l_2=50............................(ast^2)#.
#"Utilising "(ast^1) and (ast^2)" in "(ast)," we have, "#
#"The Reqd. Lim.="e/50#.
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