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)
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Limit of ((6+5*x)/(-1+5*x))^((1+2*x^2)/x)
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Limit of (-3*5^(1+x)+5*2^x)/(2*5^x+100*2^x)
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Limit of (4^x-9^x)/(4^(1+x)+9^(1+x))
- Integral of d{x}:
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x/(4+x^2)
- Graphing y =:
- x/(4+x^2)
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Identical expressions
- x/(four +x^ two)
- x divide by (4 plus x squared )
- x divide by (four plus x to the power of two)
- x/(4+x2)
- x/4+x2
- x/(4+x²)
- x/(4+x to the power of 2)
- x/4+x^2
- x divide by (4+x^2)
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Similar expressions
- (4-x)/(4+x^2-5*x)
- (5+x^2+6*x)/(4+x^2+5*x)
- (2+x^2-3*x)/(4+x^2-4*x)
- x/(4-x^2)
- (8+x^2-5*x)/(4+x^2-x)
- (5+x^4+6*x)/(4+x^2)
Limit of the function x/(4+x^2)
The solution
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/ x \
lim |------|
x->2*I+| 2|
\4 + x /
$$\lim_{x \to 2 i^+}\left(\frac{x}{x^{2} + 4}\right)$$
Limit(x/(4 + x^2), x, 2*i)
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
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 2 i^-}\left(\frac{x}{x^{2} + 4}\right) = \infty$$
More at x→2*i from the left
$$\lim_{x \to 2 i^+}\left(\frac{x}{x^{2} + 4}\right) = \infty$$
$$\lim_{x \to \infty}\left(\frac{x}{x^{2} + 4}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x}{x^{2} + 4}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x}{x^{2} + 4}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x}{x^{2} + 4}\right) = \frac{1}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x}{x^{2} + 4}\right) = \frac{1}{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x}{x^{2} + 4}\right) = 0$$
More at x→-oo
More at x→2*i from the left
$$\lim_{x \to 2 i^+}\left(\frac{x}{x^{2} + 4}\right) = \infty$$
$$\lim_{x \to \infty}\left(\frac{x}{x^{2} + 4}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x}{x^{2} + 4}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x}{x^{2} + 4}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x}{x^{2} + 4}\right) = \frac{1}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x}{x^{2} + 4}\right) = \frac{1}{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x}{x^{2} + 4}\right) = 0$$
More at x→-oo