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(1-3/x)^x

Limit of the function (1-3/x)^x

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            x
     /    3\ 
 lim |1 - -| 
x->oo\    x/ 
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{x}$$
Limit((1 - 3/x)^x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{x}$$
transform
do replacement
$$u = \frac{x}{-3}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 3 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 3 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-3}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-3} = e^{-3}$$

The final answer:
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{x} = e^{-3}$$
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
Rapid solution [src]
 -3
e  
$$e^{-3}$$
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{x} = e^{-3}$$
$$\lim_{x \to 0^-} \left(1 - \frac{3}{x}\right)^{x} = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(1 - \frac{3}{x}\right)^{x} = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(1 - \frac{3}{x}\right)^{x} = -2$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(1 - \frac{3}{x}\right)^{x} = -2$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(1 - \frac{3}{x}\right)^{x} = e^{-3}$$
More at x→-oo
The graph
Limit of the function (1-3/x)^x