Mister Exam
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Limit of the function
:
Limit of n*(1+(1+n)^2)/((1+n)*(1+n^2))
Limit of -2+|-2+x|/x
Limit of (1+x^2+9*x)/(-5+2*x+7*x^2)
Limit of (1+x^2)^(3/2)/(1+2*x)
Derivative of
:
cos(t)
Integral of d{x}
:
cos(t)
Identical expressions
cos(t)
co sinus of e of (t)
cost
Limit of the function
/
cos(t)
Limit of the function cos(t)
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 cos(t) t->oo
lim
t
→
∞
cos
(
t
)
\lim_{t \to \infty} \cos{\left(t \right)}
t
→
∞
lim
cos
(
t
)
Limit(cos(t), t, 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
2
-2
Plot the graph
Rapid solution
[src]
<-1, 1>
⟨
−
1
,
1
⟩
\left\langle -1, 1\right\rangle
⟨
−
1
,
1
⟩
Expand and simplify
Other limits t→0, -oo, +oo, 1
lim
t
→
∞
cos
(
t
)
=
⟨
−
1
,
1
⟩
\lim_{t \to \infty} \cos{\left(t \right)} = \left\langle -1, 1\right\rangle
t
→
∞
lim
cos
(
t
)
=
⟨
−
1
,
1
⟩
lim
t
→
0
−
cos
(
t
)
=
1
\lim_{t \to 0^-} \cos{\left(t \right)} = 1
t
→
0
−
lim
cos
(
t
)
=
1
More at t→0 from the left
lim
t
→
0
+
cos
(
t
)
=
1
\lim_{t \to 0^+} \cos{\left(t \right)} = 1
t
→
0
+
lim
cos
(
t
)
=
1
More at t→0 from the right
lim
t
→
1
−
cos
(
t
)
=
cos
(
1
)
\lim_{t \to 1^-} \cos{\left(t \right)} = \cos{\left(1 \right)}
t
→
1
−
lim
cos
(
t
)
=
cos
(
1
)
More at t→1 from the left
lim
t
→
1
+
cos
(
t
)
=
cos
(
1
)
\lim_{t \to 1^+} \cos{\left(t \right)} = \cos{\left(1 \right)}
t
→
1
+
lim
cos
(
t
)
=
cos
(
1
)
More at t→1 from the right
lim
t
→
−
∞
cos
(
t
)
=
⟨
−
1
,
1
⟩
\lim_{t \to -\infty} \cos{\left(t \right)} = \left\langle -1, 1\right\rangle
t
→
−
∞
lim
cos
(
t
)
=
⟨
−
1
,
1
⟩
More at t→-oo
The graph