Mister Exam
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How to use it?
- Limit of the function:
-
Limit of (10+x^2-7*x)/(8-x^3)
-
Limit of (-3-5*x+2*x^2)/(-9+x^2)
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Limit of (1+x^2)^(2/x)
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Limit of (-9-x)/(-2+x)
-
Identical expressions
- (- nine -x)/(- two +x)
- ( minus 9 minus x) divide by ( minus 2 plus x)
- ( minus nine minus x) divide by ( minus two plus x)
- -9-x/-2+x
- (-9-x) divide by (-2+x)
-
Similar expressions
- (-9+x)/(-2+x)
- (-9-x)/(-2-x)
- (-9-x)/(2+x)
- (9-x)/(-2+x)
Limit of the function (-9-x)/(-2+x)
The solution
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/-9 - x\ lim |------| x->oo\-2 + x/
$$\lim_{x \to \infty}\left(\frac{- x - 9}{x - 2}\right)$$
Limit((-9 - x)/(-2 + x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{- x - 9}{x - 2}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{- x - 9}{x - 2}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{-1 - \frac{9}{x}}{1 - \frac{2}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{-1 - \frac{9}{x}}{1 - \frac{2}{x}}\right) = \lim_{u \to 0^+}\left(\frac{- 9 u - 1}{1 - 2 u}\right)$$
=
$$\frac{-1 - 0}{1 - 0} = -1$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{- x - 9}{x - 2}\right) = -1$$
$$\lim_{x \to \infty}\left(\frac{- x - 9}{x - 2}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{- x - 9}{x - 2}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{-1 - \frac{9}{x}}{1 - \frac{2}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{-1 - \frac{9}{x}}{1 - \frac{2}{x}}\right) = \lim_{u \to 0^+}\left(\frac{- 9 u - 1}{1 - 2 u}\right)$$
=
$$\frac{-1 - 0}{1 - 0} = -1$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{- x - 9}{x - 2}\right) = -1$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(- x - 9\right) = -\infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x - 2\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{- x - 9}{x - 2}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(- x - 9\right)}{\frac{d}{d x} \left(x - 2\right)}\right)$$
=
$$\lim_{x \to \infty} -1$$
=
$$\lim_{x \to \infty} -1$$
=
$$-1$$
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
$$\lim_{x \to \infty}\left(- x - 9\right) = -\infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x - 2\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{- x - 9}{x - 2}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(- x - 9\right)}{\frac{d}{d x} \left(x - 2\right)}\right)$$
=
$$\lim_{x \to \infty} -1$$
=
$$\lim_{x \to \infty} -1$$
=
$$-1$$
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
$$\lim_{x \to \infty}\left(\frac{- x - 9}{x - 2}\right) = -1$$
$$\lim_{x \to 0^-}\left(\frac{- x - 9}{x - 2}\right) = \frac{9}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{- x - 9}{x - 2}\right) = \frac{9}{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{- x - 9}{x - 2}\right) = 10$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{- x - 9}{x - 2}\right) = 10$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{- x - 9}{x - 2}\right) = -1$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{- x - 9}{x - 2}\right) = \frac{9}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{- x - 9}{x - 2}\right) = \frac{9}{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{- x - 9}{x - 2}\right) = 10$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{- x - 9}{x - 2}\right) = 10$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{- x - 9}{x - 2}\right) = -1$$
More at x→-oo
The graph