find L- Hospital
lim x_---- inf 1/e^(-x)=
You don't need L'Hopital's rule for that limit.
1/e^(-x) = e^x
and that goes to infinity as x-> infinity.
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limxexsitat2 -4x+3/xsquer-1
To find the limit of the function \(f(x) = \frac{1}{e^{-x}}\) as \(x\) approaches infinity, we can use L'Hôpital's Rule. L'Hôpital's Rule states that for certain indeterminate forms, if the limit of the ratio of the derivatives of two functions exists, then the limit of the original functions also exists and is equal to the same value.
In our case, we have an indeterminate form of the type \(\frac{\infty}{\infty}\) when \(x\) approaches infinity. We can apply L'Hôpital's Rule by taking the derivative of the numerator and denominator with respect to \(x\) and evaluating the limit again. Let's go through the steps:
\(f(x) = \frac{1}{e^{-x}}\)
Taking the derivative of the numerator:
\(f'(x) = 0\) (since the derivative of a constant is zero)
Taking the derivative of the denominator:
\(g'(x) = (-1) \cdot e^{-x}\) (using the chain rule)
Now, we can compute the limit of the ratio of the derivatives:
\[\lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{0}{(-1) \cdot e^{-x}} = 0\]
Since the limit of the ratio of the derivatives exists and is equal to 0, we can conclude that the limit of \(f(x)\) as \(x\) approaches infinity is also 0:
\[\lim_{x \to \infty} \frac{1}{e^{-x}} = 0\]