Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (x+x^2)/(-6+x^2-5*x)
-
Limit of ((2+x)/(-5+x))^(8+3*x)
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Limit of (x^2+x^3)/tan(x^2)
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Limit of -1+x^2/(-3+x)
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Identical expressions
- - one +x^ two /(- three +x)
- minus 1 plus x squared divide by ( minus 3 plus x)
- minus one plus x to the power of two divide by ( minus three plus x)
- -1+x2/(-3+x)
- -1+x2/-3+x
- -1+x²/(-3+x)
- -1+x to the power of 2/(-3+x)
- -1+x^2/-3+x
- -1+x^2 divide by (-3+x)
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Similar expressions
- 1+x^2/(-3+x)
- -1+x^2/(3+x)
- -1+x^2/(-3-x)
- -1-x^2/(-3+x)
Limit of the function -1+x^2/(-3+x)
The solution
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/ 2 \
| x |
lim |-1 + ------|
x->oo\ -3 + x/
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x - 3} - 1\right)$$
Limit(-1 + x^2/(-3 + x), x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(x^{2} - x + 3\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x - 3\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x - 3} - 1\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{x^{2} - x + 3}{x - 3}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x^{2} - x + 3\right)}{\frac{d}{d x} \left(x - 3\right)}\right)$$
=
$$\lim_{x \to \infty}\left(2 x - 1\right)$$
=
$$\lim_{x \to \infty}\left(2 x - 1\right)$$
=
$$\infty$$
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
$$\lim_{x \to \infty}\left(x^{2} - x + 3\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x - 3\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x - 3} - 1\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{x^{2} - x + 3}{x - 3}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x^{2} - x + 3\right)}{\frac{d}{d x} \left(x - 3\right)}\right)$$
=
$$\lim_{x \to \infty}\left(2 x - 1\right)$$
=
$$\lim_{x \to \infty}\left(2 x - 1\right)$$
=
$$\infty$$
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
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x - 3} - 1\right) = \infty$$
$$\lim_{x \to 0^-}\left(\frac{x^{2}}{x - 3} - 1\right) = -1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{2}}{x - 3} - 1\right) = -1$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{2}}{x - 3} - 1\right) = - \frac{3}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{2}}{x - 3} - 1\right) = - \frac{3}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x - 3} - 1\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{x^{2}}{x - 3} - 1\right) = -1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{2}}{x - 3} - 1\right) = -1$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{2}}{x - 3} - 1\right) = - \frac{3}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{2}}{x - 3} - 1\right) = - \frac{3}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x - 3} - 1\right) = -\infty$$
More at x→-oo
The graph