Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (-2+x^2-x)/(-2-3*x+2*x^2)
-
Limit of (-20+x^2-x)/(12+x^2+7*x)
-
Limit of ((1+x+x^2)/(x^2+3*x))^(-5+4*x)
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Limit of 1+2/x+3*x
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Identical expressions
- one + two /x+ three *x
- 1 plus 2 divide by x plus 3 multiply by x
- one plus two divide by x plus three multiply by x
- 1+2/x+3x
- 1+2 divide by x+3*x
-
Similar expressions
- 1-2/x+3*x
- 1+2/x-3*x
Limit of the function 1+2/x+3*x
The solution
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/ 2 \ lim |1 + - + 3*x| x->oo\ x /
$$\lim_{x \to \infty}\left(3 x + \left(1 + \frac{2}{x}\right)\right)$$
Limit(1 + 2/x + 3*x, x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(3 x^{2} + x + 2\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} x = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(3 x + \left(1 + \frac{2}{x}\right)\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{3 x^{2} + x + 2}{x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(3 x^{2} + x + 2\right)}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to \infty}\left(6 x + 1\right)$$
=
$$\lim_{x \to \infty}\left(6 x + 1\right)$$
=
$$\infty$$
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
$$\lim_{x \to \infty}\left(3 x^{2} + x + 2\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} x = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(3 x + \left(1 + \frac{2}{x}\right)\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{3 x^{2} + x + 2}{x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(3 x^{2} + x + 2\right)}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to \infty}\left(6 x + 1\right)$$
=
$$\lim_{x \to \infty}\left(6 x + 1\right)$$
=
$$\infty$$
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
$$\lim_{x \to \infty}\left(3 x + \left(1 + \frac{2}{x}\right)\right) = \infty$$
$$\lim_{x \to 0^-}\left(3 x + \left(1 + \frac{2}{x}\right)\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(3 x + \left(1 + \frac{2}{x}\right)\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(3 x + \left(1 + \frac{2}{x}\right)\right) = 6$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(3 x + \left(1 + \frac{2}{x}\right)\right) = 6$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(3 x + \left(1 + \frac{2}{x}\right)\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(3 x + \left(1 + \frac{2}{x}\right)\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(3 x + \left(1 + \frac{2}{x}\right)\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(3 x + \left(1 + \frac{2}{x}\right)\right) = 6$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(3 x + \left(1 + \frac{2}{x}\right)\right) = 6$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(3 x + \left(1 + \frac{2}{x}\right)\right) = -\infty$$
More at x→-oo
The graph