lim𝑥→0𝑒−(1+2𝑥)12𝑥𝑥 is equal to

The given limit is of the form 0/0 when x tends to 0, so we can apply L'Hopital's rule.

L'Hopital's rule states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.

So, we need to find the derivatives of the numerator and the denominator.

The derivative of the numerator, e^(-1/(1+2x)), using the chain rule, is e^(-1/(1+2x)) * (1/(1+2x)^2) * (-2).

The derivative of the denominator, x, is 1.

So, the limit becomes:

lim (x->0) [e^(-1/(1+2x)) * (1/(1+2x)^2) * (-2)] / 1

This simplifies to:

lim (x->0) -2e^(-1/(1+2x)) / (1+2x)^2

As x tends to 0, the expression -1/(1+2x) tends to -1, so e^(-1/(1+2x)) tends to e^-1 or 1/e.

The denominator (1+2x)^2 tends to 1.

So, the limit is -2(1/e) / 1 = -2/e.

To solve this limit, we can use L'Hopital's Rule, which states that the limit as x approaches a of [f(x) - g(x)] / [h(x) - k(x)] is equal to the limit as x approaches a of [f'(x) / h'(x)].

Here, we can rewrite the expression as a fraction:

lim(x->0) [e^(-1/(1+2x)) - 1] / x

Now, we can apply L'Hopital's Rule. We need to find the derivative of the numerator and the derivative of the denominator.

The derivative of the numerator, using the chain rule, is:

e^(-1/(1+2x)) * (1/(1+2x)^2) * (-2)

The derivative of the denominator is simply 1.

So, we have:

lim(x->0) { [e^(-1/(1+2x)) * (1/(1+2x)^2) * (-2)] / 1 }

As x approaches 0, the expression simplifies to:

-2 * e^(-1) * 1 = -2/e

So, the limit of the given expression as x approaches 0 is -2/e.