Math Final

Let’s solve Question 1 step by step, explaining each limit evaluation clearly.

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Question 1: Limits

a. ${\lim }_{x\to +\infty }\frac{x^3+2x^2-x}{x^4-5x}$

Step 1: Analyze the dominant terms
For large $x$, the dominant term in the numerator is $x^3$, and in the denominator, it is $x^4$. The other terms become negligible as $x\to \infty$.

Step 2: Divide numerator and denominator by the highest power of $x$ in the denominator ($x^4$):

$\frac{x^3+2x^2-x}{x^4-5x}=\frac{\frac{x^3}{x^4}+\frac{2x^2}{x^4}-\frac{x}{x^4}}{\frac{x^4}{x^4}-\frac{5x}{x^4}}=\frac{\frac{1}{x}+\frac{2}{x^2}-\frac{1}{x^3}}{1-\frac{5}{x^3}}.$

Step 3: Simplify as $x\to +\infty$:

$\frac{\frac{1}{x}+\frac{2}{x^2}-\frac{1}{x^3}}{1-\frac{5}{x^3}}\to \frac{0+0-0}{1-0}=0.$

Final Answer:

${\lim }_{x\to +\infty }\frac{x^3+2x^2-x}{x^4-5x}=0.$

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b. ${\lim }_{x\to 3^{+}}\frac{2x-6}{\ln (x^2-8)}.$

Step 1: Analyze the numerator and denominator as $(x\to 3^{+})$:

• Numerator: $(2x-6)approaches(2(3)-6=0).$
• Denominator: $(\ln (x^2-8))$ approaches $(\ln (3^2-8)=\ln (1)=0).$

This results in an indeterminate form $(\frac{0}{0}).$

Step 2: Apply L’Hôpital’s Rule:

Since we have an indeterminate form, we can apply L’Hôpital’s Rule, which involves differentiating the numerator and the denominator:

• Derivative of the numerator: $(\frac{d}{dx}(2x-6)=2)$.
• Derivative of the denominator: $(\frac{d}{dx}(\ln (x^2-8))=\frac{1}{x^2-8}\cdot 2x=\frac{2x}{x^2-8})$.

Step 3: Evaluate the limit using the derivatives:

${\lim }_{x\to 3^{+}}\frac{2}{\frac{2x}{x^2-8}}={\lim }_{x\to 3^{+}}\frac{2(x^2-8)}{2x}={\lim }_{x\to 3^{+}}\frac{x^2-8}{x}.$

Simplify the expression:

${\lim }_{x\to 3^{+}}\frac{x^2-8}{x}={\lim }_{x\to 3^{+}}\left(x-\frac{8}{x}\right).$

Step 4: Substitute (x = 3):

${\lim }_{x\to 3^{+}}\left(x-\frac{8}{x}\right)=3-\frac{8}{3}=\frac{9}{3}-\frac{8}{3}=\frac{1}{3}.$

Final Answer:

The limit as $(x)$ approaches 3 from the right is $(\frac{1}{3}).$

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${\lim }_{x\to -2}\frac{x-2}{x^2+2x}$

1. Substitute $x=-2$:
   • Numerator: $x-2=-2-2=-4$.
   • Denominator: $x^2+2x=(-2{)}^2+2(-2)=4-4=0$.

This results in an indeterminate form $\frac{-4}{0}$, which suggests that the limit might not exist in the traditional sense. However, let’s simplify the expression to see if we can resolve it:

2. Factor the Denominator:

   • The denominator $x^2+2x$ can be factored as $x(x+2)$.

3. Simplify the Expression:

   • The expression becomes $\frac{x-2}{x(x+2)}$.

4. Evaluate the Limit:

   • Since the denominator becomes zero at $x=-2$, and the numerator does not, the limit approaches negative or positive infinity depending on the direction from which $x$ approaches $-2$.

Thus, the limit ${\lim }_{x\to -2}\frac{x-2}{x^2+2x}$ does not exist in the traditional sense, as the expression approaches infinity. To determine the behavior more precisely, you would need to consider the one-sided limits:

• As $x$ approaches $-2$ from the left ($x\to -2^{-}$), the expression approaches $-\infty$.
• As $x$ approaches $-2$ from the right ($x\to -2^{+}$), the expression approaches $+\infty$.

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d. ${\lim }_{x\to 0^{+}}(5x+\cos x{)}^{\frac{3}{x}}$

1. Base Analysis:

   • As $x\to 0^{+}$, $5x\to 0$ and $\cos x\to \cos (0)=1$.
   • Therefore, $5x+\cos x\to 0+1=1$.

2. Exponent Analysis:

   • The exponent $\frac{3}{x}$ approaches $+\infty$ as $x\to 0^{+}$.

3. Overall Expression:

   • We have an expression of the form $1^{\infty }$, which is an indeterminate form.

To resolve this indeterminate form, we can use the natural logarithm and L’Hôpital’s Rule:

4. Take the Natural Logarithm:

   • Let $y=(5x+\cos x{)}^{\frac{3}{x}}$.
   • Then $\ln y=\frac{3}{x}\ln (5x+\cos x)$.

5. Evaluate the Limit of $\ln y$:

   • Consider ${\lim }_{x\to 0^{+}}\frac{3}{x}\ln (5x+\cos x)$.
   • As $x\to 0^{+}$, $\ln (5x+\cos x)\to \ln (1)=0$.
   • We have an indeterminate form $\frac{0}{0}$, so we can apply L’Hôpital’s Rule.

6. Apply L’Hôpital’s Rule:

   • Differentiate the numerator and the denominator:
     • Derivative of the numerator: $\frac{d}{dx}[\ln (5x+\cos x)]=\frac{5-\sin x}{5x+\cos x}$.
     • Derivative of the denominator: $\frac{d}{dx}[x]=1$.
   • The limit becomes ${\lim }_{x\to 0^{+}}3\cdot \frac{5-\sin x}{x(5x+\cos x)}$.

7. Evaluate the Limit:

   • As $x\to 0^{+}$, $\frac{5-\sin x}{5x+\cos x}\to \frac{5-0}{0+1}=5$.
   • Therefore, the limit of $\ln y$ is ${\lim }_{x\to 0^{+}}3\cdot 5=15$.

8. Exponentiate to Find $y$:

   • Since $\ln y\to 15$, $y\to e^{15}$.

Final Answer: The limit is $e^{15}$.

Given Piecewise Function:

$$f(x)=\{\begin{array}{ll}\frac{x-2}{\sqrt{x+7}-3},& x>2,\\ 5x+b,& x=2,\\ 3+ax,& 0<x<2,\\ 3+e^{2x},& x\le 0.\end{array}$$

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a. Is (f(x)) continuous at (x = 0)? Justify.

To check continuity at (x = 0), the following must hold:

1. $(f(0))$ is defined.
2. $({\lim }_{x\to 0^{-}}f(x)$ = ${\lim }_{x\to 0^{+}}f(x)=f(0))$.

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Step 1: Evaluate (f(0)):

For $(x\le 0),(f(x)=3+e^{2x})$. Substituting $(x=0)$:

$$f(0)=3+e^0=3+1=4.$$

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Step 2: Evaluate $({\lim }_{x\to 0^{-}}f(x)):$

For $@$ As $(x\to 0^{-}):$

$$\underset{x\to 0^{-}}{\lim }f(x)=3+e^0=4.$$

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Step 3: Evaluate: $({\lim }_{x\to 0^{+}}f(x))$

For $(0<x<2),(f(x)=3+ax).$ As $(x\to 0^{+}):$

$$\underset{x\to 0^{+}}{\lim }f(x)=3+a(0)=3.$$

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Step 4: Compare limits and (f(0)):

$$\underset{x\to 0^{-}}{\lim }f(x)=4,\underset{x\to 0^{+}}{\lim }f(x)=3,f(0)=4.$$

Since $({\lim }_{x\to 0^{-}}f(x)\ne {\lim }_{x\to 0^{+}}f(x))$, (f(x)) is not continuous at (x = 0).

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b. Find the values of (a) and (b) to make (f(x)) continuous at (x = 2).

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To ensure continuity at (x = 2), the following must hold:

1. ${\lim }_{x\to 2^{-}}f(x)=f(2)$,
2. ${\lim }_{x\to 2^{+}}f(x)=f(2)$.

Step 1: Compute (f(2)):
At (x = 2):

$$f(2)=5(2)+b=10+b.$$

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**Step 2: Compute (\lim_{x \to 2^-} f(x)) $For$(0 < x < 2), (f(x) = 3 + ax).$As$(x \to 2

\lim_{x \to 2^-} f(x) = 3 + a(2) = 3 + 2a. $$ \] --- **Step 3: Compute \(\lim_{x \to 2^+} f(x)\):** For \(x > 2\), $$(f(x) = frac{x - 2}{\sqrt{x + 7} - 3}). As (x \to 2^+):$$ 1. Simplify the denominator: $$[ \sqrt{x + 7} - 3. ]$$ At $$(x = 2), (\sqrt{x + 7} = \sqrt{9} = 3)$$, so direct substitution leads to an indeterminate form $$(\frac{0}{0}).$$ 2. Use L'Hôpital's Rule: Differentiate the numerator and denominator: - Numerator: $$(\frac{d}{dx}[x - 2] = 1),$$ - Denominator: \(\frac{d}{dx}[\sqrt{x + 7} - 3] = \frac{1}{2\sqrt{x + 7}}\). Applying L'Hôpital's Rule: \[ $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2} \frac{1}{\frac{1}{2\sqrt{x + 7}}} = \lim_{x \to 2} 2\sqrt{x + 7}.

Substituting (x = 2):
[
${\lim }_{x\to 2^{+}}f(x)=2\sqrt{9}=6.$
]

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