Mister Exam
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How to use it?
- Limit of the function:
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Limit of (-3+sqrt(1+4*x))/(-2+x)
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Limit of 1/(-8+x)
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Limit of (1/x)^(log(1+x)/log(2-x))
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Limit of (1-cos(6*x))/(1-cos(8*x))
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Identical expressions
- one /(- eight +x)
- 1 divide by ( minus 8 plus x)
- one divide by ( minus eight plus x)
- 1/-8+x
- 1 divide by (-8+x)
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Similar expressions
- 1/(8+x)
- 1/(-8-x)
Limit of the function 1/(-8+x)
The solution
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[src]
1 lim ------ x->8+-8 + x
$$\lim_{x \to 8^+} \frac{1}{x - 8}$$
Limit(1/(-8 + x), x, 8)
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
$$\lim_{x \to 8^-} \frac{1}{x - 8} = \infty$$
More at x→8 from the left
$$\lim_{x \to 8^+} \frac{1}{x - 8} = \infty$$
$$\lim_{x \to \infty} \frac{1}{x - 8} = 0$$
More at x→oo
$$\lim_{x \to 0^-} \frac{1}{x - 8} = - \frac{1}{8}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \frac{1}{x - 8} = - \frac{1}{8}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \frac{1}{x - 8} = - \frac{1}{7}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \frac{1}{x - 8} = - \frac{1}{7}$$
More at x→1 from the right
$$\lim_{x \to -\infty} \frac{1}{x - 8} = 0$$
More at x→-oo
More at x→8 from the left
$$\lim_{x \to 8^+} \frac{1}{x - 8} = \infty$$
$$\lim_{x \to \infty} \frac{1}{x - 8} = 0$$
More at x→oo
$$\lim_{x \to 0^-} \frac{1}{x - 8} = - \frac{1}{8}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \frac{1}{x - 8} = - \frac{1}{8}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \frac{1}{x - 8} = - \frac{1}{7}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \frac{1}{x - 8} = - \frac{1}{7}$$
More at x→1 from the right
$$\lim_{x \to -\infty} \frac{1}{x - 8} = 0$$
More at x→-oo
One‐sided limits
[src]
1 lim ------ x->8+-8 + x
$$\lim_{x \to 8^+} \frac{1}{x - 8}$$
oo
$$\infty$$
= 151.0
1 lim ------ x->8--8 + x
$$\lim_{x \to 8^-} \frac{1}{x - 8}$$
-oo
$$-\infty$$
= -151.0
= -151.0
The graph