Calculus/L'Hôpital's rule/Solutions
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Calculus
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L'Hôpital's rule
L'Hôpital's rule Solutions
1.
lim
x
→
0
x
+
tan
x
sin
x
{\displaystyle \lim _{x\to 0}{\frac {x+\tan x}{\sin x}}}
lim
x
→
0
1
+
sec
2
x
cos
x
=
2
{\displaystyle \lim _{x\to 0}{\frac {1+\sec ^{2}x}{\cos x}}=\mathbf {2} }
2.
lim
x
→
π
x
−
π
sin
x
{\displaystyle \lim _{x\to \pi }{\frac {x-\pi }{\sin x}}}
lim
x
→
π
1
cos
x
=
−
1
{\displaystyle \lim _{x\to \pi }{\frac {1}{\cos x}}=\mathbf {-1} }
3.
lim
x
→
0
sin
3
x
sin
4
x
{\displaystyle \lim _{x\to 0}{\frac {\sin 3x}{\sin 4x}}}
lim
x
→
0
3
cos
(
3
x
)
4
cos
(
4
x
)
=
3
4
{\displaystyle \lim _{x\to 0}{\frac {3\cos(3x)}{4\cos(4x)}}=\mathbf {\frac {3}{4}} }
4.
lim
x
→
∞
x
5
e
5
x
{\displaystyle \lim _{x\to \infty }{\frac {x^{5}}{e^{5x}}}}
lim
x
→
∞
5
x
4
5
e
5
x
=
lim
x
→
∞
5
⋅
4
x
3
5
2
e
5
x
=
lim
x
→
∞
5
⋅
4
⋅
3
x
2
5
3
e
5
x
=
lim
x
→
∞
5
⋅
4
⋅
3
⋅
2
x
5
4
e
5
x
=
lim
x
→
∞
5
⋅
4
⋅
3
⋅
2
⋅
1
5
5
e
5
x
=
0
{\displaystyle {\begin{aligned}\lim _{x\to \infty }{\frac {5x^{4}}{5e^{5x}}}&=\lim _{x\to \infty }{\frac {5\cdot 4x^{3}}{5^{2}e^{5x}}}\\&=\lim _{x\to \infty }{\frac {5\cdot 4\cdot 3x^{2}}{5^{3}e^{5x}}}\\&=\lim _{x\to \infty }{\frac {5\cdot 4\cdot 3\cdot 2x}{5^{4}e^{5x}}}\\&=\lim _{x\to \infty }{\frac {5\cdot 4\cdot 3\cdot 2\cdot 1}{5^{5}e^{5x}}}\\&=\mathbf {0} \end{aligned}}}
5.
lim
x
→
0
tan
x
−
x
sin
x
−
x
{\displaystyle \lim _{x\to 0}{\frac {\tan x-x}{\sin x-x}}}
lim
x
→
0
sec
2
x
−
1
cos
x
−
1
=
lim
x
→
0
2
sec
x
cos
−
2
x
sin
x
−
sin
x
=
−
2
{\displaystyle \lim _{x\to 0}{\frac {\sec ^{2}x-1}{\cos x-1}}=\lim _{x\to 0}{\frac {2\sec x\cos ^{-2}x\sin x}{-\sin x}}=\mathbf {-2} }
