Mister Exam
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How to use it?
- Limit of the function:
- Limit of (1/factorial(x))^(1/x)
- Limit of (1+cos(3*x))^(1/cos(x))
-
Limit of (1-cos(2*x))/(1-cos(4*x))
-
Limit of (1-4*x)/(-2+5*x)
-
Identical expressions
- (one - four *x)/(- two + five *x)
- (1 minus 4 multiply by x) divide by ( minus 2 plus 5 multiply by x)
- (one minus four multiply by x) divide by ( minus two plus five multiply by x)
- (1-4x)/(-2+5x)
- 1-4x/-2+5x
- (1-4*x) divide by (-2+5*x)
-
Similar expressions
- (1-4*x)/(2+5*x)
- (1+4*x)/(-2+5*x)
- (1-4*x)/(-2-5*x)
Limit of the function (1-4*x)/(-2+5*x)
The solution
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/1 - 4*x \ lim |--------| x->oo\-2 + 5*x/
$$\lim_{x \to \infty}\left(\frac{1 - 4 x}{5 x - 2}\right)$$
Limit((1 - 4*x)/(-2 + 5*x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{1 - 4 x}{5 x - 2}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{1 - 4 x}{5 x - 2}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{-4 + \frac{1}{x}}{5 - \frac{2}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{-4 + \frac{1}{x}}{5 - \frac{2}{x}}\right) = \lim_{u \to 0^+}\left(\frac{u - 4}{5 - 2 u}\right)$$
=
$$\frac{-4}{5 - 0} = - \frac{4}{5}$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{1 - 4 x}{5 x - 2}\right) = - \frac{4}{5}$$
$$\lim_{x \to \infty}\left(\frac{1 - 4 x}{5 x - 2}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{1 - 4 x}{5 x - 2}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{-4 + \frac{1}{x}}{5 - \frac{2}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{-4 + \frac{1}{x}}{5 - \frac{2}{x}}\right) = \lim_{u \to 0^+}\left(\frac{u - 4}{5 - 2 u}\right)$$
=
$$\frac{-4}{5 - 0} = - \frac{4}{5}$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{1 - 4 x}{5 x - 2}\right) = - \frac{4}{5}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(1 - 4 x\right) = -\infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(5 x - 2\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{1 - 4 x}{5 x - 2}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(1 - 4 x\right)}{\frac{d}{d x} \left(5 x - 2\right)}\right)$$
=
$$\lim_{x \to \infty} - \frac{4}{5}$$
=
$$\lim_{x \to \infty} - \frac{4}{5}$$
=
$$- \frac{4}{5}$$
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)
-oo/oo,
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(1 - 4 x\right) = -\infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(5 x - 2\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{1 - 4 x}{5 x - 2}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(1 - 4 x\right)}{\frac{d}{d x} \left(5 x - 2\right)}\right)$$
=
$$\lim_{x \to \infty} - \frac{4}{5}$$
=
$$\lim_{x \to \infty} - \frac{4}{5}$$
=
$$- \frac{4}{5}$$
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
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(\frac{1 - 4 x}{5 x - 2}\right) = - \frac{4}{5}$$
$$\lim_{x \to 0^-}\left(\frac{1 - 4 x}{5 x - 2}\right) = - \frac{1}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{1 - 4 x}{5 x - 2}\right) = - \frac{1}{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{1 - 4 x}{5 x - 2}\right) = -1$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{1 - 4 x}{5 x - 2}\right) = -1$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{1 - 4 x}{5 x - 2}\right) = - \frac{4}{5}$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{1 - 4 x}{5 x - 2}\right) = - \frac{1}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{1 - 4 x}{5 x - 2}\right) = - \frac{1}{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{1 - 4 x}{5 x - 2}\right) = -1$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{1 - 4 x}{5 x - 2}\right) = -1$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{1 - 4 x}{5 x - 2}\right) = - \frac{4}{5}$$
More at x→-oo
The graph