Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
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Limit of 1-cos(3*x)
-
Limit of (-10+6*x^2+7*x)/(-4+x^2)
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Limit of (7+n)/(5+n)
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Limit of -3+x*(7-6*x)^x/3
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Identical expressions
- five /(- nine +x+x^ four)
- 5 divide by ( minus 9 plus x plus x to the power of 4)
- five divide by ( minus nine plus x plus x to the power of four)
- 5/(-9+x+x4)
- 5/-9+x+x4
- 5/(-9+x+x⁴)
- 5/-9+x+x^4
- 5 divide by (-9+x+x^4)
-
Similar expressions
- 5/(-9-x+x^4)
- 5/(9+x+x^4)
- 5/(-9+x-x^4)
Limit of the function 5/(-9+x+x^4)
The solution
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/ 5 \
lim |-----------|
x->oo| 4|
\-9 + x + x /
$$\lim_{x \to \infty}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right)$$
Limit(5/(-9 + x + x^4), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right)$$
Let's divide numerator and denominator by x^4:
$$\lim_{x \to \infty}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{5 \frac{1}{x^{4}}}{1 + \frac{1}{x^{3}} - \frac{9}{x^{4}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{5 \frac{1}{x^{4}}}{1 + \frac{1}{x^{3}} - \frac{9}{x^{4}}}\right) = \lim_{u \to 0^+}\left(\frac{5 u^{4}}{- 9 u^{4} + u^{3} + 1}\right)$$
=
$$\frac{5 \cdot 0^{4}}{0^{3} - 9 \cdot 0^{4} + 1} = 0$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = 0$$
$$\lim_{x \to \infty}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right)$$
Let's divide numerator and denominator by x^4:
$$\lim_{x \to \infty}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{5 \frac{1}{x^{4}}}{1 + \frac{1}{x^{3}} - \frac{9}{x^{4}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{5 \frac{1}{x^{4}}}{1 + \frac{1}{x^{3}} - \frac{9}{x^{4}}}\right) = \lim_{u \to 0^+}\left(\frac{5 u^{4}}{- 9 u^{4} + u^{3} + 1}\right)$$
=
$$\frac{5 \cdot 0^{4}}{0^{3} - 9 \cdot 0^{4} + 1} = 0$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = 0$$
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = 0$$
$$\lim_{x \to 0^-}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = - \frac{5}{9}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = - \frac{5}{9}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = - \frac{5}{7}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = - \frac{5}{7}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = 0$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = - \frac{5}{9}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = - \frac{5}{9}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = - \frac{5}{7}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = - \frac{5}{7}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{5}{x^{4} + \left(x - 9\right)}\right) = 0$$
More at x→-oo
The graph