Question

Lim((1-cosx)/(e^(2x)-e^(x)) x-->0

Answer 1

Ответ:

0

Пошаговое объяснение:

Придел числителя lim(-cos(x)+1)=0

Придел знаменателя lim(e^2x-e^x)=0

Потом надо найти все производные пока не избавимся от неопределенности

lim((e^-x/e^x-1)*(-cosx+1))

=

lim(sinx/2e^2x-e^x)

Answer 2

To find the limit of the given expression as x approaches 0, we can use L'Hôpital's rule, which allows us to evaluate limits of indeterminate forms. An indeterminate form is when both the numerator and denominator of a fraction approach zero (or infinity) as x approaches the limiting value.

The expression we need to evaluate is:

Lim(x --> 0) [(1 - cosx) / (e^(2x) - e^(x))]

Since both the numerator and denominator approach 0 as x approaches 0, we can apply L'Hôpital's rule by taking the derivative of the numerator and denominator with respect to x.

Step 1: Find the derivative of the numerator and denominator.

Numerator: d/dx (1 - cosx) = 0 + sinx = sinx

Denominator: d/dx (e^(2x) - e^(x)) = 2e^(2x) - e^(x)

Step 2: Evaluate the limit of the derivatives as x approaches 0.

Lim(x --> 0) sinx = sin(0) = 0

Lim(x --> 0) (2e^(2x) - e^(x)) = 2e^(2*0) - e^(0) = 2 - 1 = 1

Step 3: Take the ratio of the limits of the derivatives to find the final limit.

Lim(x --> 0) [(1 - cosx) / (e^(2x) - e^(x))] = Lim(x --> 0) [sinx / (2e^(2x) - e^(x))]

= (Lim(x --> 0) sinx) / (Lim(x --> 0) (2e^(2x) - e^(x)))

= 0 / 1

= 0

So, the limit of the given expression as x approaches 0 is 0.