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Sagot :
Certainly! Let's evaluate the limit of the given expression as [tex]\( x \)[/tex] approaches [tex]\( 0 \)[/tex] from the left. The expression is:
[tex]\[ \lim _{x \rightarrow 0^{-}}\left(\left[\frac{\sin x}{x}\right]+\left[\frac{\sin 3x}{x}\right]+\left[\frac{5 \sin^{-1} (4x)}{x}\right]+\left[\frac{10 \tan^{-1} (2x)}{3x}\right]\right) \][/tex]
To find this limit, let's analyze each component of the expression as [tex]\( x \)[/tex] approaches 0.
1. First Component:
[tex]\[ \lim_{x \to 0^{-}} \frac{\sin x}{x} \][/tex]
It is well-known that:
[tex]\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \][/tex]
2. Second Component:
[tex]\[ \lim_{x \to 0^{-}} \frac{\sin 3x}{x} \][/tex]
Using the substitution [tex]\( y = 3x \)[/tex]:
[tex]\[ \lim_{x \to 0^{-}} \frac{\sin 3x}{x} = \lim_{y \to 0^{-}} \frac{\sin y}{y/3} = 3 \lim_{y \to 0^{-}} \frac{\sin y}{y} = 3 \][/tex]
3. Third Component:
[tex]\[ \lim_{x \to 0^{-}} \frac{5 \sin^{-1} (4x)}{x} \][/tex]
Using the substitution [tex]\( z = 4x \)[/tex]:
[tex]\[ \lim_{x \to 0^{-}} \frac{5 \sin^{-1} (4x)}{x} = 5 \lim_{z \to 0^{-}} \frac{\sin^{-1} z}{z/4} = 20 \lim_{z \to 0^{-}} \frac{\sin^{-1} z}{z} = 20 \][/tex]
Here, we use the fact that:
[tex]\[ \lim_{z \to 0} \frac{\sin^{-1} z}{z} = 1 \][/tex]
4. Fourth Component:
[tex]\[ \lim_{x \to 0^{-}} \frac{10 \tan^{-1} (2x)}{3x} \][/tex]
Using the substitution [tex]\( w = 2x \)[/tex]:
[tex]\[ \lim_{x \to 0^{-}} \frac{10 \tan^{-1} (2x)}{3x} = \frac{10}{3} \lim_{w \to 0^{-}} \frac{\tan^{-1} w}{w/2} = \frac{20}{3} \lim_{w \to 0^{-}} \frac{\tan^{-1} w}{w} = \frac{20}{3} \][/tex]
Here, we use the fact that:
[tex]\[ \lim_{w \to 0} \frac{\tan^{-1} w}{w} = 1 \][/tex]
Combining all these results, the limit of the original expression is:
[tex]\[ \lim_{x \to 0^{-}} \left( \frac{\sin x}{x} + \frac{\sin 3x}{x} + \frac{5 \sin^{-1}(4x)}{x} + \frac{10 \tan^{-1}(2x)}{3x} \right) = 1 + 3 + 20 + \frac{20}{3} \][/tex]
Now, adding these:
[tex]\[ 1 + 3 + 20 + \frac{20}{3} = \frac{3}{3} + \frac{9}{3} + \frac{60}{3} + \frac{20}{3} = \frac{3 + 9 + 60 + 20}{3} = \frac{92}{3} \][/tex]
Therefore, the solution is:
[tex]\[ \lim _{x \rightarrow 0^{-}}\left(\left[\frac{\sin x}{x}\right]+\left[\frac{\sin 3 x}{x}\right]+\left[\frac{5 \sin ^{-1} 4 x}{x}\right]+\left[\frac{10 \tan ^{-1} 2 x}{3 x}\right]\right) = \frac{92}{3} \][/tex]
[tex]\[ \lim _{x \rightarrow 0^{-}}\left(\left[\frac{\sin x}{x}\right]+\left[\frac{\sin 3x}{x}\right]+\left[\frac{5 \sin^{-1} (4x)}{x}\right]+\left[\frac{10 \tan^{-1} (2x)}{3x}\right]\right) \][/tex]
To find this limit, let's analyze each component of the expression as [tex]\( x \)[/tex] approaches 0.
1. First Component:
[tex]\[ \lim_{x \to 0^{-}} \frac{\sin x}{x} \][/tex]
It is well-known that:
[tex]\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \][/tex]
2. Second Component:
[tex]\[ \lim_{x \to 0^{-}} \frac{\sin 3x}{x} \][/tex]
Using the substitution [tex]\( y = 3x \)[/tex]:
[tex]\[ \lim_{x \to 0^{-}} \frac{\sin 3x}{x} = \lim_{y \to 0^{-}} \frac{\sin y}{y/3} = 3 \lim_{y \to 0^{-}} \frac{\sin y}{y} = 3 \][/tex]
3. Third Component:
[tex]\[ \lim_{x \to 0^{-}} \frac{5 \sin^{-1} (4x)}{x} \][/tex]
Using the substitution [tex]\( z = 4x \)[/tex]:
[tex]\[ \lim_{x \to 0^{-}} \frac{5 \sin^{-1} (4x)}{x} = 5 \lim_{z \to 0^{-}} \frac{\sin^{-1} z}{z/4} = 20 \lim_{z \to 0^{-}} \frac{\sin^{-1} z}{z} = 20 \][/tex]
Here, we use the fact that:
[tex]\[ \lim_{z \to 0} \frac{\sin^{-1} z}{z} = 1 \][/tex]
4. Fourth Component:
[tex]\[ \lim_{x \to 0^{-}} \frac{10 \tan^{-1} (2x)}{3x} \][/tex]
Using the substitution [tex]\( w = 2x \)[/tex]:
[tex]\[ \lim_{x \to 0^{-}} \frac{10 \tan^{-1} (2x)}{3x} = \frac{10}{3} \lim_{w \to 0^{-}} \frac{\tan^{-1} w}{w/2} = \frac{20}{3} \lim_{w \to 0^{-}} \frac{\tan^{-1} w}{w} = \frac{20}{3} \][/tex]
Here, we use the fact that:
[tex]\[ \lim_{w \to 0} \frac{\tan^{-1} w}{w} = 1 \][/tex]
Combining all these results, the limit of the original expression is:
[tex]\[ \lim_{x \to 0^{-}} \left( \frac{\sin x}{x} + \frac{\sin 3x}{x} + \frac{5 \sin^{-1}(4x)}{x} + \frac{10 \tan^{-1}(2x)}{3x} \right) = 1 + 3 + 20 + \frac{20}{3} \][/tex]
Now, adding these:
[tex]\[ 1 + 3 + 20 + \frac{20}{3} = \frac{3}{3} + \frac{9}{3} + \frac{60}{3} + \frac{20}{3} = \frac{3 + 9 + 60 + 20}{3} = \frac{92}{3} \][/tex]
Therefore, the solution is:
[tex]\[ \lim _{x \rightarrow 0^{-}}\left(\left[\frac{\sin x}{x}\right]+\left[\frac{\sin 3 x}{x}\right]+\left[\frac{5 \sin ^{-1} 4 x}{x}\right]+\left[\frac{10 \tan ^{-1} 2 x}{3 x}\right]\right) = \frac{92}{3} \][/tex]
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