Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (1-4*x)^(1/x)
-
Limit of (-16+x^2+6*x)/(-2-5*x+3*x^2)
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Limit of (1+x)^(2/3)-(-1+x)^(2/3)
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Limit of 1/3+x/3
- Derivative of:
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x^12
- Graphing y =:
- x^12
- Integral of d{x}:
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x^12
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Identical expressions
- x^ twelve
- x to the power of 12
- x to the power of twelve
- x12
Limit of the function x^12
The solution
Detail solution
Let's take the limit
$$\lim_{x \to \infty} x^{12}$$
Let's divide numerator and denominator by x^12:
$$\lim_{x \to \infty} x^{12}$$ =
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x^{12}}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x^{12}}} = \lim_{u \to 0^+} \frac{1}{u^{12}}$$
=
$$\frac{1}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty} x^{12} = \infty$$
$$\lim_{x \to \infty} x^{12}$$
Let's divide numerator and denominator by x^12:
$$\lim_{x \to \infty} x^{12}$$ =
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x^{12}}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x^{12}}} = \lim_{u \to 0^+} \frac{1}{u^{12}}$$
=
$$\frac{1}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty} x^{12} = \infty$$
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
$$\lim_{x \to \infty} x^{12} = \infty$$
$$\lim_{x \to 0^-} x^{12} = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+} x^{12} = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-} x^{12} = 1$$
More at x→1 from the left
$$\lim_{x \to 1^+} x^{12} = 1$$
More at x→1 from the right
$$\lim_{x \to -\infty} x^{12} = \infty$$
More at x→-oo
$$\lim_{x \to 0^-} x^{12} = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+} x^{12} = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-} x^{12} = 1$$
More at x→1 from the left
$$\lim_{x \to 1^+} x^{12} = 1$$
More at x→1 from the right
$$\lim_{x \to -\infty} x^{12} = \infty$$
More at x→-oo
The graph