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Sagot :
Sure! Let's go through each limit step-by-step.
### i. [tex]\(\lim_{{x \to -\infty}} \left(2 + \frac{1}{x}\right)^3\)[/tex]
1. As [tex]\(x \rightarrow -\infty\)[/tex], the term [tex]\(\frac{1}{x}\)[/tex] approaches 0 because [tex]\(\frac{1}{x}\)[/tex] gets smaller and smaller.
2. Therefore, [tex]\(2 + \frac{1}{x}\)[/tex] approaches 2 since the [tex]\(\frac{1}{x}\)[/tex] term becomes negligible.
3. Raising this result to the power of 3, we get:
[tex]\[ \left(2 + \frac{1}{x}\right)^3 \to 2^3 = 8 \][/tex]
4. Thus,
[tex]\[ \lim_{{x \to -\infty}} \left(2 + \frac{1}{x}\right)^3 = 8 \][/tex]
### ii. [tex]\(\lim_{{n \to \infty}} \left(x^4 - 7x^2 + 3x - 5\right)\)[/tex]
1. This expression is a polynomial in [tex]\(x\)[/tex] and not dependent on [tex]\(n\)[/tex].
2. Therefore, the limit of this expression as [tex]\(n \rightarrow \infty\)[/tex] is just the expression itself, since it does not change with [tex]\(n\)[/tex]:
[tex]\[ \lim_{{n \to \infty}} \left(x^4 - 7x^2 + 3x - 5\right) = x^4 - 7x^2 + 3x - 5 \][/tex]
### iii. [tex]\(\lim_{{x \to 2}} \left(\frac{4 - x^2}{3 - \sqrt{x^2 + 5}}\right)\)[/tex]
1. First, substitute [tex]\(x = 2\)[/tex] directly into the function where possible:
[tex]\[ 4 - x^2 \text{ at } x = 2 \rightarrow 4 - 2^2 = 4 - 4 = 0 \][/tex]
[tex]\[ \sqrt{x^2 + 5} \text{ at } x = 2 \rightarrow \sqrt{2^2 + 5} = \sqrt{4 + 5} = \sqrt{9} = 3 \][/tex]
2. Thus, the expression becomes:
[tex]\[ \frac{4 - 2^2}{3 - \sqrt{2^2 + 5}} = \frac{0}{3 - 3} = \frac{0}{0} \][/tex]
This is an indeterminate form, so we need another approach.
3. To properly resolve the limit, we can apply L'Hôpital's Rule which states that if we have an indeterminate form of [tex]\(\frac{0}{0}\)[/tex], we can take the derivatives of the numerator and the denominator:
[tex]\[ \text{Numerator: } \frac{d}{dx} (4 - x^2) = -2x \][/tex]
[tex]\[ \text{Denominator: } \frac{d}{dx} (3 - \sqrt{x^2 + 5}) = -\frac{x}{\sqrt{x^2 + 5}} \][/tex]
Applying L'Hôpital's Rule:
[tex]\[ \lim_{{x \to 2}} \frac{-2x}{-\frac{x}{\sqrt{x^2 + 5}}} = \lim_{{x \to 2}} \frac{-2x \cdot \sqrt{x^2 + 5}}{-x} \][/tex]
4. Simplify:
[tex]\[ \lim_{{x \to 2}} \frac{2x \cdot \sqrt{x^2 + 5}}{x} = \lim_{{x \to 2}} 2 \sqrt{x^2 + 5} \][/tex]
Substitute [tex]\(x = 2\)[/tex] back in:
[tex]\[ 2 \sqrt{2^2 + 5} = 2 \sqrt{4 + 5} = 2 \cdot 3 = 6 \][/tex]
5. Therefore,
[tex]\[ \lim_{{x \to 2}} \left(\frac{4 - x^2}{3 - \sqrt{x^2 + 5}}\right) = 6 \][/tex]
We have successfully evaluated all the given limits:
1. [tex]\(\lim_{{x \to -\infty}} \left(2 + \frac{1}{x}\right)^3 = 8\)[/tex]
2. [tex]\(\lim_{{n \to \infty}} \left(x^4 - 7x^2 + 3x - 5\right) = x^4 - 7x^2 + 3x - 5\)[/tex]
3. [tex]\(\lim_{{x \to 2}} \left(\frac{4 - x^2}{3 - \sqrt{x^2 + 5}}\right) = 6\)[/tex]
### i. [tex]\(\lim_{{x \to -\infty}} \left(2 + \frac{1}{x}\right)^3\)[/tex]
1. As [tex]\(x \rightarrow -\infty\)[/tex], the term [tex]\(\frac{1}{x}\)[/tex] approaches 0 because [tex]\(\frac{1}{x}\)[/tex] gets smaller and smaller.
2. Therefore, [tex]\(2 + \frac{1}{x}\)[/tex] approaches 2 since the [tex]\(\frac{1}{x}\)[/tex] term becomes negligible.
3. Raising this result to the power of 3, we get:
[tex]\[ \left(2 + \frac{1}{x}\right)^3 \to 2^3 = 8 \][/tex]
4. Thus,
[tex]\[ \lim_{{x \to -\infty}} \left(2 + \frac{1}{x}\right)^3 = 8 \][/tex]
### ii. [tex]\(\lim_{{n \to \infty}} \left(x^4 - 7x^2 + 3x - 5\right)\)[/tex]
1. This expression is a polynomial in [tex]\(x\)[/tex] and not dependent on [tex]\(n\)[/tex].
2. Therefore, the limit of this expression as [tex]\(n \rightarrow \infty\)[/tex] is just the expression itself, since it does not change with [tex]\(n\)[/tex]:
[tex]\[ \lim_{{n \to \infty}} \left(x^4 - 7x^2 + 3x - 5\right) = x^4 - 7x^2 + 3x - 5 \][/tex]
### iii. [tex]\(\lim_{{x \to 2}} \left(\frac{4 - x^2}{3 - \sqrt{x^2 + 5}}\right)\)[/tex]
1. First, substitute [tex]\(x = 2\)[/tex] directly into the function where possible:
[tex]\[ 4 - x^2 \text{ at } x = 2 \rightarrow 4 - 2^2 = 4 - 4 = 0 \][/tex]
[tex]\[ \sqrt{x^2 + 5} \text{ at } x = 2 \rightarrow \sqrt{2^2 + 5} = \sqrt{4 + 5} = \sqrt{9} = 3 \][/tex]
2. Thus, the expression becomes:
[tex]\[ \frac{4 - 2^2}{3 - \sqrt{2^2 + 5}} = \frac{0}{3 - 3} = \frac{0}{0} \][/tex]
This is an indeterminate form, so we need another approach.
3. To properly resolve the limit, we can apply L'Hôpital's Rule which states that if we have an indeterminate form of [tex]\(\frac{0}{0}\)[/tex], we can take the derivatives of the numerator and the denominator:
[tex]\[ \text{Numerator: } \frac{d}{dx} (4 - x^2) = -2x \][/tex]
[tex]\[ \text{Denominator: } \frac{d}{dx} (3 - \sqrt{x^2 + 5}) = -\frac{x}{\sqrt{x^2 + 5}} \][/tex]
Applying L'Hôpital's Rule:
[tex]\[ \lim_{{x \to 2}} \frac{-2x}{-\frac{x}{\sqrt{x^2 + 5}}} = \lim_{{x \to 2}} \frac{-2x \cdot \sqrt{x^2 + 5}}{-x} \][/tex]
4. Simplify:
[tex]\[ \lim_{{x \to 2}} \frac{2x \cdot \sqrt{x^2 + 5}}{x} = \lim_{{x \to 2}} 2 \sqrt{x^2 + 5} \][/tex]
Substitute [tex]\(x = 2\)[/tex] back in:
[tex]\[ 2 \sqrt{2^2 + 5} = 2 \sqrt{4 + 5} = 2 \cdot 3 = 6 \][/tex]
5. Therefore,
[tex]\[ \lim_{{x \to 2}} \left(\frac{4 - x^2}{3 - \sqrt{x^2 + 5}}\right) = 6 \][/tex]
We have successfully evaluated all the given limits:
1. [tex]\(\lim_{{x \to -\infty}} \left(2 + \frac{1}{x}\right)^3 = 8\)[/tex]
2. [tex]\(\lim_{{n \to \infty}} \left(x^4 - 7x^2 + 3x - 5\right) = x^4 - 7x^2 + 3x - 5\)[/tex]
3. [tex]\(\lim_{{x \to 2}} \left(\frac{4 - x^2}{3 - \sqrt{x^2 + 5}}\right) = 6\)[/tex]
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