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Evaluate the limit, lim_x⭢ 0(sin x+x+2)(cos x+1).
Related Topics
Wize University Calculus 1 Textbook > Limits
Limit Laws
2 Activities
Evaluate the limit,
lim
x
→
0
(
sin
x
+
x
+
2
)
(
cos
x
+
1
)
\lim_{x\rightarrow 0}(\sin x+x+2)(\cos x+1)
lim
x
→
0
(
sin
x
+
x
+
2
)
(
cos
x
+
1
)
.
Answer
I don't know
Check Submission
More Limit Laws Questions:
Limit Laws
lim
x
→
π
2
cos
x
−
π
sin
x
=
\displaystyle \lim_{x\to\frac{\pi}{2}}\ \frac{\cos x-\pi}{\sin x}=
x
→
2
π
lim
sin
x
cos
x
−
π
=
Limit with Absolute Value
lim
x
→
0
∣
3
−
2
x
∣
−
∣
2
x
−
3
∣
x
=
\displaystyle\lim_{x\to0}\ \frac{\left|3-2x\right|-\left|2x-3\right|}{x}=
x
→
0
lim
x
∣
3
−
2
x
∣
−
∣
2
x
−
3
∣
=
Limit Laws
If
lim
x
→
∞
cos
(
arctan
(
x
2
−
2
)
)
f
(
x
)
=
π
\displaystyle \lim_{x\to\infty}\frac{\cos\left(\arctan\left(x^2-2\right)\right)}{f\left(x\right)}=\pi
x
→
∞
lim
f
(
x
)
cos
(
arctan
(
x
2
−
2
)
)
=
π
, then
lim
x
→
∞
cos
(
f
(
x
)
)
\displaystyle \lim_{x\to\infty}\cos\left(f\left(x\right)\right)
x
→
∞
lim
cos
(
f
(
x
)
)
is
Limit Laws
lim
x
→
π
2
cos
x
−
π
sin
x
=
\displaystyle \lim_{x\to\frac{\pi}{2}}\ \frac{\cos x-\pi}{\sin x}=
x
→
2
π
lim
sin
x
cos
x
−
π
=
Basic limit techniques - Absolute value
Evaluate
lim
x
→
4
−
∣
x
−
4
∣
x
2
−
4
.
\lim\limits_{x\rightarrow4^-}\frac{\vert x-4\vert}{x^2-4}.
x
→
4
−
lim
x
2
−
4
∣
x
−
4∣
.
Limit Laws
lim
x
→
π
2
cos
x
−
π
sin
x
=
\displaystyle \lim_{x\to\frac{\pi}{2}}\ \frac{\cos x-\pi}{\sin x}=
x
→
2
π
lim
sin
x
cos
x
−
π
=
Basic limit techniques - Absolute value
Evaluate
lim
x
→
4
−
∣
x
−
4
∣
x
2
−
4
.
\lim\limits_{x\rightarrow4^-}\frac{\vert x-4\vert}{x^2-4}.
x
→
4
−
lim
x
2
−
4
∣
x
−
4∣
.
Evaluating limits
Determine the limit
lim
x
→
0
∣
5
x
−
2
∣
−
∣
5
x
+
2
∣
2
x
\lim\ _{x\rightarrow0\ }\frac{\left|5x-2\right|-\left|5x+2\right|}{2x}
lim
x
→
0
2
x
∣
5
x
−
2
∣
−
∣
5
x
+
2
∣
Limit Law & Direct Substitution
If
lim
x
→
π
f
(
x
)
sin
2
x
=
−
1
3
\displaystyle \lim_{x\to\pi}\ \frac{f\left(x\right)}{\sin^2x}=-\frac{1}{3}
x
→
π
lim
sin
2
x
f
(
x
)
=
−
3
1
, then what is the value of
lim
x
→
π
f
(
x
)
sin
x
\displaystyle \lim_{x\to\pi}\ \frac{f\left(x\right)}{\sin x}
x
→
π
lim
sin
x
f
(
x
)
?
Limit Laws
If
lim
x
→
∞
cos
(
arctan
(
x
2
−
2
)
)
f
(
x
)
=
π
\displaystyle \lim_{x\to\infty}\frac{\cos\left(\arctan\left(x^2-2\right)\right)}{f\left(x\right)}=\pi
x
→
∞
lim
f
(
x
)
cos
(
arctan
(
x
2
−
2
)
)
=
π
, then
lim
x
→
∞
cos
(
f
(
x
)
)
\displaystyle \lim_{x\to\infty}\cos\left(f\left(x\right)\right)
x
→
∞
lim
cos
(
f
(
x
)
)
is
Limit Laws
lim
x
→
π
2
cos
x
−
π
sin
x
=
\displaystyle \lim_{x\to\frac{\pi}{2}}\ \frac{\cos x-\pi}{\sin x}=
x
→
2
π
lim
sin
x
cos
x
−
π
=
Basic limit techniques - Absolute value
Evaluate
lim
x
→
4
−
∣
x
−
4
∣
x
2
−
4
.
\lim\limits_{x\rightarrow4^-}\frac{\vert x-4\vert}{x^2-4}.
x
→
4
−
lim
x
2
−
4
∣
x
−
4∣
.
Determine the value of
lim
x
→
0
(
4
+
x
)
2
−
16
x
\displaystyle \lim_{x\rightarrow 0}\frac{(4+x)^2-16}{x}
x
→
0
lim
x
(
4
+
x
)
2
−
16
.
Determine the value of
lim
x
→
(
π
/
2
)
+
sin
2
(
x
)
−
x
sin
x
\displaystyle \lim_{x\rightarrow(\pi/2)^+}\frac{\sin^2 (x) -x}{\sin x}
x
→
(
π
/2
)
+
lim
sin
x
sin
2
(
x
)
−
x
.
Evaluate the limit,
lim
x
→
2
(
x
2
+
2
x
+
1
)
\lim_{x\rightarrow 2}(x^2+2x+1)
lim
x
→
2
(
x
2
+
2
x
+
1
)
.
Limit Law & Direct Substitution
If
lim
x
→
π
f
(
x
)
sin
2
x
=
−
1
3
\displaystyle \lim_{x\to\pi}\ \frac{f\left(x\right)}{\sin^2x}=-\frac{1}{3}
x
→
π
lim
sin
2
x
f
(
x
)
=
−
3
1
, then what is the value of
lim
x
→
π
f
(
x
)
sin
x
\displaystyle \lim_{x\to\pi}\ \frac{f\left(x\right)}{\sin x}
x
→
π
lim
sin
x
f
(
x
)
?
Find the value of b such that
lim
x
→
0
a
x
+
b
−
5
x
=
2
\lim_{x\rightarrow0}\frac{\sqrt{ax+b-5}}{x}=2
lim
x
→
0
x
a
x
+
b
−
5
=
2
.
Limits: Indeterminate Forms
If
lim
x
→
2
4
−
2
f
(
x
)
2
−
x
\displaystyle\lim_{x\ \rightarrow2}\ \frac{4-2\ f\left(x\right)}{2-x}
x
→
2
lim
2
−
x
4
−
2
f
(
x
)
is defined, what is the value of
lim
x
→
2
f
(
x
)
\displaystyle\lim_{x\ \rightarrow2}\ \ f\left(x\right)
x
→
2
lim
f
(
x
)
?
Evaluating limits
Determine the limit
lim
x
→
0
∣
5
x
−
2
∣
−
∣
5
x
+
2
∣
2
x
\lim\ _{x\rightarrow0\ }\frac{\left|5x-2\right|-\left|5x+2\right|}{2x}
lim
x
→
0
2
x
∣
5
x
−
2
∣
−
∣
5
x
+
2
∣
Limit with Absolute Value
lim
x
→
0
∣
3
−
2
x
∣
−
∣
2
x
−
3
∣
x
=
\displaystyle\lim_{x\to0}\ \frac{\left|3-2x\right|-\left|2x-3\right|}{x}=
x
→
0
lim
x
∣
3
−
2
x
∣
−
∣
2
x
−
3
∣
=
Limits: Basics
Evaluate the limit:
lim
x
→
3
[
x
x
2
−
5
+
1
]
2
\displaystyle\lim_{x\rightarrow3}\left[\frac{x}{x^2-5}+1\right]^2
x
→
3
lim
[
x
2
−
5
x
+
1
]
2