Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of x^(4/x)
-
Limit of (-6+x^2-x)/(-4+x^2)
-
Limit of (-6+x^2-x)/(-21+x+2*x^2)
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Limit of e^(3-x)*(-2+x)
- Derivative of:
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16/x
- Graphing y =:
- 16/x
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Identical expressions
- sixteen /x
- 16 divide by x
- sixteen divide by x
Limit of the function 16/x
The solution
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/16\ lim |--| x->oo\x /
$$\lim_{x \to \infty}\left(\frac{16}{x}\right)$$
Limit(16/x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{16}{x}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{16}{x}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{16 \frac{1}{x}}{1}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{16 \frac{1}{x}}{1}\right) = \lim_{u \to 0^+}\left(16 u\right)$$
=
$$0 \cdot 16 = 0$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{16}{x}\right) = 0$$
$$\lim_{x \to \infty}\left(\frac{16}{x}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{16}{x}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{16 \frac{1}{x}}{1}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{16 \frac{1}{x}}{1}\right) = \lim_{u \to 0^+}\left(16 u\right)$$
=
$$0 \cdot 16 = 0$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{16}{x}\right) = 0$$
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
$$\lim_{x \to \infty}\left(\frac{16}{x}\right) = 0$$
$$\lim_{x \to 0^-}\left(\frac{16}{x}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{16}{x}\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{16}{x}\right) = 16$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{16}{x}\right) = 16$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{16}{x}\right) = 0$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{16}{x}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{16}{x}\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{16}{x}\right) = 16$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{16}{x}\right) = 16$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{16}{x}\right) = 0$$
More at x→-oo
The graph