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How to use it?
Limit of the function
:
Limit of ((3+x)^2+(3-x)^2)/((3-x)^2-(3+x)^2)
Limit of (1-log(7*x))^(7*x)
Limit of (-2+x^2-x)/(-2+x)
Limit of (1-3*x)^(1/x)
Derivative of
:
log(1-x)
Graphing y =
:
log(1-x)
Integral of d{x}
:
log(1-x)
Identical expressions
log(one -x)
logarithm of (1 minus x)
logarithm of (one minus x)
log1-x
Similar expressions
log(1-x^2)/(pi/2-acot(x))
log(1+x)
Limit of the function
/
log(1-x)
Limit of the function log(1-x)
at
→
Calculate the limit!
v
For end points:
---------
From the left (x0-)
From the right (x0+)
The graph:
from
to
Piecewise:
{
enter the piecewise function here
The solution
You have entered
[src]
lim log(1 - x) x->oo
lim
x
→
∞
log
(
−
x
+
1
)
\lim_{x \to \infty} \log{\left(- x + 1 \right)}
x
→
∞
lim
lo
g
(
−
x
+
1
)
Limit(log(1 - x), x, oo, dir='-')
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
0
2
4
6
8
-8
-6
-4
-2
-10
10
5
-5
Plot the graph
Rapid solution
[src]
oo
∞
\infty
∞
Expand and simplify
Other limits x→0, -oo, +oo, 1
lim
x
→
∞
log
(
−
x
+
1
)
=
∞
\lim_{x \to \infty} \log{\left(- x + 1 \right)} = \infty
x
→
∞
lim
lo
g
(
−
x
+
1
)
=
∞
lim
x
→
0
−
log
(
−
x
+
1
)
=
0
\lim_{x \to 0^-} \log{\left(- x + 1 \right)} = 0
x
→
0
−
lim
lo
g
(
−
x
+
1
)
=
0
More at x→0 from the left
lim
x
→
0
+
log
(
−
x
+
1
)
=
0
\lim_{x \to 0^+} \log{\left(- x + 1 \right)} = 0
x
→
0
+
lim
lo
g
(
−
x
+
1
)
=
0
More at x→0 from the right
lim
x
→
1
−
log
(
−
x
+
1
)
=
−
∞
\lim_{x \to 1^-} \log{\left(- x + 1 \right)} = -\infty
x
→
1
−
lim
lo
g
(
−
x
+
1
)
=
−
∞
More at x→1 from the left
lim
x
→
1
+
log
(
−
x
+
1
)
=
−
∞
\lim_{x \to 1^+} \log{\left(- x + 1 \right)} = -\infty
x
→
1
+
lim
lo
g
(
−
x
+
1
)
=
−
∞
More at x→1 from the right
lim
x
→
−
∞
log
(
−
x
+
1
)
=
∞
\lim_{x \to -\infty} \log{\left(- x + 1 \right)} = \infty
x
→
−
∞
lim
lo
g
(
−
x
+
1
)
=
∞
More at x→-oo
The graph