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# what is Lim x tends to 0 ( 1 +ax) to the power of 1/x ?

Nandana
110 Points
4 years ago
it’s become 1
as lim x → 0 (1+a*0) = 1
& 1 ^ (infinite ) = 1
so the answer is  ‘1’
this is also done by expanding of  Binomial  series, then also we get 1
by this way -
1+ 1/x(ax) /1! + 1/x(1/(x-1))/2! + …
as we seen that except 1st term all terms are multiplied by ‘x’, as x tending to ‘0’ these terms also =0
& we left with ‘1’ .

Nandana
110 Points
4 years ago
sorry , it’s my mistake it’s not ‘1’ , it is e^a ,
lim  x → 0 (1 +ax )^(1/x)  is obtained by expansion of binomial series as follows –
=>   (01/x)(1*(ax)^0) + (11/x)(ax)^1 + (21/x)(ax)^2 + …. (1/x1/x) (ax)^(1/x)
=>  1+ (a/1!)+(a^2/2!)+ (a^3/3!)+ …..
=> e^a  [since e^x = 1+(x/1!)+(x^2/2!)+(x^3/3!)+ … ]