Question

(a) Consider the function f(x)=(e^x −1)/x.

Use l’Hˆopital’s rule to show that lim f(x) = 1 when x approaches 0

(b) Check this result empirically by writing a program to compute f(x) for x = 10−k, k = 1,...,15. Do your results agree with theoretical expectations? Explain why.

(c) Perform the experiment in part b again, this time using the mathematically equivalent formulation, f(x)=(e^x −1)/log(e^x), evaluated as indicated, with no simplification. If this works any better, can you explain why?

Please ignore the programming part. I am interested in knowing why this happens.

Answer 1

By L'H^opital's rule, in any limit value

it in indeterminate form so, that is 0/0, here we can apply  L'H^opital's rule, so that we get it in determined limit value.

By this rule we can differentiate f(x) and g(x) separately, and then it is in ratio we can evaluate. if still it is same that is indeterminate form then again we can differentiate. So this sequence goes on until we get the finite value.

Let's see how we can do this differentiation, by rule derivation please see/refer

Now differentiation of f(x) = e^x-1 is f'(x)=e^x-0=e^x, and differentiation of g(x) = x is g'(x)=1

so substitute this value in lim and we get lim x>0 e^x/1=e^0/1=1. Hence it is proved.

It is happens because limit value of e^x-1 nearer to 0 same as limit value of x nearer to 0, as we know limit value not substitute value. so it tends to same value and we get 1. Take a value based on substitute x=0.0001, so we get e^0.0001 is=1.00010000 and minus 1 we get = 0.0001 and denominator we have 0.0001, so both value same and ratio we calculate is 1. By this way we can get based on calculation as well. For negative value we can check as well, let's take x= - 0.0001, then e^ - 0.0001 = 0.99990000 then minus 1 we get - 0.0001 so divide by x value that is -0.0001 , both ratio same so we get ration is 1. By this calculation we get always value of ratio tending towards 1, so we can say limit value of ratio of function when x tends to 0, always tends to 1.