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