Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of x^2*(1/3-cos(8*x)/3)
-
Limit of (-2+x^2-x)/(-2+x+3*x^2)
-
Limit of (-1+sin(x))/cos(x)
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Limit of (-2*sin(x)+sin(2*x))/(x*log(cos(5*x)))
- Integral of d{x}:
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1/(4+x^2)
- Sum of series:
- 1/(4+x^2)
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Identical expressions
- one /(four +x^ two)
- 1 divide by (4 plus x squared )
- one divide by (four plus x to the power of two)
- 1/(4+x2)
- 1/4+x2
- 1/(4+x²)
- 1/(4+x to the power of 2)
- 1/4+x^2
- 1 divide by (4+x^2)
-
Similar expressions
- (-1+8*x^3)/(-1/4+x^2)
- 1/(4-x^2)
Limit of the function 1/(4+x^2)
The solution
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1
lim ------
x->-oo 2
4 + x
$$\lim_{x \to -\infty} \frac{1}{x^{2} + 4}$$
Limit(1/(4 + x^2), x, -oo)
Detail solution
Let's take the limit
$$\lim_{x \to -\infty} \frac{1}{x^{2} + 4}$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to -\infty} \frac{1}{x^{2} + 4}$$ =
$$\lim_{x \to -\infty}\left(\frac{1}{x^{2} \left(1 + \frac{4}{x^{2}}\right)}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty}\left(\frac{1}{x^{2} \left(1 + \frac{4}{x^{2}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u^{2}}{4 u^{2} + 1}\right)$$
=
$$\frac{0^{2}}{4 \cdot 0^{2} + 1} = 0$$
The final answer:
$$\lim_{x \to -\infty} \frac{1}{x^{2} + 4} = 0$$
$$\lim_{x \to -\infty} \frac{1}{x^{2} + 4}$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to -\infty} \frac{1}{x^{2} + 4}$$ =
$$\lim_{x \to -\infty}\left(\frac{1}{x^{2} \left(1 + \frac{4}{x^{2}}\right)}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty}\left(\frac{1}{x^{2} \left(1 + \frac{4}{x^{2}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u^{2}}{4 u^{2} + 1}\right)$$
=
$$\frac{0^{2}}{4 \cdot 0^{2} + 1} = 0$$
The final answer:
$$\lim_{x \to -\infty} \frac{1}{x^{2} + 4} = 0$$
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
$$\lim_{x \to -\infty} \frac{1}{x^{2} + 4} = 0$$
$$\lim_{x \to \infty} \frac{1}{x^{2} + 4} = 0$$
More at x→oo
$$\lim_{x \to 0^-} \frac{1}{x^{2} + 4} = \frac{1}{4}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \frac{1}{x^{2} + 4} = \frac{1}{4}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \frac{1}{x^{2} + 4} = \frac{1}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \frac{1}{x^{2} + 4} = \frac{1}{5}$$
More at x→1 from the right
$$\lim_{x \to \infty} \frac{1}{x^{2} + 4} = 0$$
More at x→oo
$$\lim_{x \to 0^-} \frac{1}{x^{2} + 4} = \frac{1}{4}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \frac{1}{x^{2} + 4} = \frac{1}{4}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \frac{1}{x^{2} + 4} = \frac{1}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \frac{1}{x^{2} + 4} = \frac{1}{5}$$
More at x→1 from the right
The graph