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Sagot :
Let's solve each limit step-by-step.
### Part a
[tex]\[ \lim_{x \to 0} (3 - x) \cdot 2 \][/tex]
First, substitute [tex]\( x = 0 \)[/tex] directly into the expression:
[tex]\[ (3 - 0) \cdot 2 = 3 \cdot 2 = 6 \][/tex]
Hence,
[tex]\[ \boxed{6} \][/tex]
### Part b
[tex]\[ \lim_{x \to 3} \frac{x + 5x - 9}{2x} \][/tex]
Combine like terms in the numerator:
[tex]\[ x + 5x - 9 = 6x - 9 \][/tex]
So the expression becomes:
[tex]\[ \frac{6x - 9}{2x} \][/tex]
Now substitute [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{6(3) - 9}{2 \cdot 3} = \frac{18 - 9}{6} = \frac{9}{6} = \frac{3}{2} \][/tex]
Hence,
[tex]\[ \boxed{\frac{3}{2}} \][/tex]
### Part c
[tex]\[ \lim_{x \to 1} \frac{5x^2 - 8x + 15}{x - 3} \][/tex]
Direct substitution of [tex]\( x = 1 \)[/tex] into the denominator results in a non-zero value. Substituting [tex]\( x = 1 \)[/tex] into the numerator:
[tex]\[ 5(1)^2 - 8(1) + 15 = 5 - 8 + 15 = 12 \][/tex]
Since substituting [tex]\( x = 1 \)[/tex] directly gives:
[tex]\[ \frac{12}{-2} \][/tex]
Which equals:
[tex]\[ -6 \][/tex]
Hence,
[tex]\[ \boxed{-6} \][/tex]
### Part d
[tex]\[ \lim_{x \to 30} \frac{\sin x}{\cos x} \][/tex]
We know that [tex]\( \frac{\sin x}{\cos x} = \tan x \)[/tex]. Thus, we have:
[tex]\[ \lim_{x \to 30} \tan x \][/tex]
Substitute [tex]\( x = 30 \)[/tex]:
[tex]\[ \tan 30 = \frac{\sin 30}{\cos 30} \][/tex]
Hence,
[tex]\[ \boxed{\frac{\sin 30}{\cos 30}} \][/tex]
### Part e
[tex]\[ \lim_{x \to 1} \frac{25x^3 + 12x^2 + 9x - 28}{50x^2 + 13x} \][/tex]
Substitute [tex]\( x = 1 \)[/tex] in both the numerator and the denominator directly:
Numerator:
[tex]\[ 25(1)^3 + 12(1)^2 + 9(1) - 28 = 25 + 12 + 9 - 28 = 18 \][/tex]
Denominator:
[tex]\[ 50(1)^2 + 13(1) = 50 + 13 = 63 \][/tex]
So the expression becomes:
[tex]\[ \frac{18}{63} = \frac{2}{7} \][/tex]
Hence,
[tex]\[ \boxed{\frac{2}{7}} \][/tex]
These are the solutions for the given limits.
### Part a
[tex]\[ \lim_{x \to 0} (3 - x) \cdot 2 \][/tex]
First, substitute [tex]\( x = 0 \)[/tex] directly into the expression:
[tex]\[ (3 - 0) \cdot 2 = 3 \cdot 2 = 6 \][/tex]
Hence,
[tex]\[ \boxed{6} \][/tex]
### Part b
[tex]\[ \lim_{x \to 3} \frac{x + 5x - 9}{2x} \][/tex]
Combine like terms in the numerator:
[tex]\[ x + 5x - 9 = 6x - 9 \][/tex]
So the expression becomes:
[tex]\[ \frac{6x - 9}{2x} \][/tex]
Now substitute [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{6(3) - 9}{2 \cdot 3} = \frac{18 - 9}{6} = \frac{9}{6} = \frac{3}{2} \][/tex]
Hence,
[tex]\[ \boxed{\frac{3}{2}} \][/tex]
### Part c
[tex]\[ \lim_{x \to 1} \frac{5x^2 - 8x + 15}{x - 3} \][/tex]
Direct substitution of [tex]\( x = 1 \)[/tex] into the denominator results in a non-zero value. Substituting [tex]\( x = 1 \)[/tex] into the numerator:
[tex]\[ 5(1)^2 - 8(1) + 15 = 5 - 8 + 15 = 12 \][/tex]
Since substituting [tex]\( x = 1 \)[/tex] directly gives:
[tex]\[ \frac{12}{-2} \][/tex]
Which equals:
[tex]\[ -6 \][/tex]
Hence,
[tex]\[ \boxed{-6} \][/tex]
### Part d
[tex]\[ \lim_{x \to 30} \frac{\sin x}{\cos x} \][/tex]
We know that [tex]\( \frac{\sin x}{\cos x} = \tan x \)[/tex]. Thus, we have:
[tex]\[ \lim_{x \to 30} \tan x \][/tex]
Substitute [tex]\( x = 30 \)[/tex]:
[tex]\[ \tan 30 = \frac{\sin 30}{\cos 30} \][/tex]
Hence,
[tex]\[ \boxed{\frac{\sin 30}{\cos 30}} \][/tex]
### Part e
[tex]\[ \lim_{x \to 1} \frac{25x^3 + 12x^2 + 9x - 28}{50x^2 + 13x} \][/tex]
Substitute [tex]\( x = 1 \)[/tex] in both the numerator and the denominator directly:
Numerator:
[tex]\[ 25(1)^3 + 12(1)^2 + 9(1) - 28 = 25 + 12 + 9 - 28 = 18 \][/tex]
Denominator:
[tex]\[ 50(1)^2 + 13(1) = 50 + 13 = 63 \][/tex]
So the expression becomes:
[tex]\[ \frac{18}{63} = \frac{2}{7} \][/tex]
Hence,
[tex]\[ \boxed{\frac{2}{7}} \][/tex]
These are the solutions for the given limits.
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