Is f(x) =(1+x)^1/x,when x is not equal to 0 and f(x)=e, when x=0 continuous at x=0.
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By definition: Euler's exponential function is defined as:
eⁿ = Lim x -> 0, (1 + x n)^(1/x) or,
(1 + n / x )^x
Given f(x) = (1 + x)^(1/x) when x ≠ 0
So Lim x -> 0 , f(x) = Lim x -> 0, (1 + 1* x)^(1/x)
= e¹ = e.
Also given f(0) = e.
Since the limit as x tends to 0 is same as f(0) , the function f(x) is continuous at x = 0.
==========================
eⁿ = Lim x -> 0, (1 + x n)^(1/x) or,
(1 + n / x )^x
Given f(x) = (1 + x)^(1/x) when x ≠ 0
So Lim x -> 0 , f(x) = Lim x -> 0, (1 + 1* x)^(1/x)
= e¹ = e.
Also given f(0) = e.
Since the limit as x tends to 0 is same as f(0) , the function f(x) is continuous at x = 0.
==========================
kvnmurty:
:-)
Then, Differentiate the numerator and denominator and apply limits afterwards. We get it.
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