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Limit of the function e^(3-x)*(-2+x)

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     / 3 - x         \
 lim \E     *(-2 + x)/
x->oo

\lim_{x \to \infty}\left(e^{3 - x} \left(x - 2\right)\right)

Limit(E^(3 - x)*(-2 + x), x, oo, dir='-')

Lopital's rule

We have indeterminateness of type

oo/oo,

i.e. limit for the numerator is

\lim_{x \to \infty}\left(x - 2\right) = \infty

and limit for the denominator is

\lim_{x \to \infty} e^{x - 3} = \infty

Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.

\lim_{x \to \infty}\left(e^{3 - x} \left(x - 2\right)\right)

=
Let's transform the function under the limit a few

\lim_{x \to \infty}\left(\left(x - 2\right) e^{3 - x}\right)

\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x - 2\right)}{\frac{d}{d x} e^{x - 3}}\right)

\lim_{x \to \infty}\left(e^{3} e^{- x}\right)

\lim_{x \to \infty}\left(e^{3} e^{- x}\right)

0

It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)

The graph

Plot the graph

Rapid solution [src]

0

Expand and simplify

Other limits x→0, -oo, +oo, 1

\lim_{x \to \infty}\left(e^{3 - x} \left(x - 2\right)\right) = 0

\lim_{x \to 0^-}\left(e^{3 - x} \left(x - 2\right)\right) = - 2 e^{3}

\lim_{x \to 0^+}\left(e^{3 - x} \left(x - 2\right)\right) = - 2 e^{3}

\lim_{x \to 1^-}\left(e^{3 - x} \left(x - 2\right)\right) = - e^{2}

\lim_{x \to 1^+}\left(e^{3 - x} \left(x - 2\right)\right) = - e^{2}

\lim_{x \to -\infty}\left(e^{3 - x} \left(x - 2\right)\right) = -\infty

The graph