Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (9-x+2*x^2)/(5-x)
-
Limit of ((1+x)^4-(-1+x)^4)/((1+x)^3+(-1+x)^3)
-
Limit of (-4+x^2)/(-8+x^3)
-
Limit of 2*x
-
Identical expressions
- (nine -x+ two *x^ two)/(five -x)
- (9 minus x plus 2 multiply by x squared ) divide by (5 minus x)
- (nine minus x plus two multiply by x to the power of two) divide by (five minus x)
- (9-x+2*x2)/(5-x)
- 9-x+2*x2/5-x
- (9-x+2*x²)/(5-x)
- (9-x+2*x to the power of 2)/(5-x)
- (9-x+2x^2)/(5-x)
- (9-x+2x2)/(5-x)
- 9-x+2x2/5-x
- 9-x+2x^2/5-x
- (9-x+2*x^2) divide by (5-x)
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Similar expressions
- (9-x+2*x^2)/(5+x)
- (9-x-2*x^2)/(5-x)
- (9+x+2*x^2)/(5-x)
Limit of the function (9-x+2*x^2)/(5-x)
The solution
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/ 2\
|9 - x + 2*x |
lim |------------|
x->oo\ 5 - x /
$$\lim_{x \to \infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right)$$
Limit((9 - x + 2*x^2)/(5 - x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{2 - \frac{1}{x} + \frac{9}{x^{2}}}{- \frac{1}{x} + \frac{5}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{2 - \frac{1}{x} + \frac{9}{x^{2}}}{- \frac{1}{x} + \frac{5}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{9 u^{2} - u + 2}{5 u^{2} - u}\right)$$
=
$$\frac{- 0 + 9 \cdot 0^{2} + 2}{- 0 + 5 \cdot 0^{2}} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = -\infty$$
$$\lim_{x \to \infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{2 - \frac{1}{x} + \frac{9}{x^{2}}}{- \frac{1}{x} + \frac{5}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{2 - \frac{1}{x} + \frac{9}{x^{2}}}{- \frac{1}{x} + \frac{5}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{9 u^{2} - u + 2}{5 u^{2} - u}\right)$$
=
$$\frac{- 0 + 9 \cdot 0^{2} + 2}{- 0 + 5 \cdot 0^{2}} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = -\infty$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(2 x^{2} - x + 9\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(5 - x\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{2 x^{2} - x + 9}{5 - x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(2 x^{2} - x + 9\right)}{\frac{d}{d x} \left(5 - x\right)}\right)$$
=
$$\lim_{x \to \infty}\left(1 - 4 x\right)$$
=
$$\lim_{x \to \infty}\left(1 - 4 x\right)$$
=
$$-\infty$$
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
$$\lim_{x \to \infty}\left(2 x^{2} - x + 9\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(5 - x\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{2 x^{2} - x + 9}{5 - x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(2 x^{2} - x + 9\right)}{\frac{d}{d x} \left(5 - x\right)}\right)$$
=
$$\lim_{x \to \infty}\left(1 - 4 x\right)$$
=
$$\lim_{x \to \infty}\left(1 - 4 x\right)$$
=
$$-\infty$$
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
$$\lim_{x \to \infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = -\infty$$
$$\lim_{x \to 0^-}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = \frac{9}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = \frac{9}{5}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = \frac{5}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = \frac{5}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = \infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = \frac{9}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = \frac{9}{5}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = \frac{5}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = \frac{5}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{2 x^{2} + \left(9 - x\right)}{5 - x}\right) = \infty$$
More at x→-oo
The graph