\begin{tabular}{l} Question $$ \mathbf{5} $$ \ \begin{tabular}{ll} 5.1 Find the limit of each of the functions given below. & $$ \mathbf{2 0} $$ Marks \ a. $$ \lim _{h ightarrow 0} \frac{(h-1)^{3}+1}{h} $$ & (4 Marks) \ b. $$ \lim _{x ightarrow \infty} \frac{4 x^{4}-3 x^{2}+1}{2 x^{4}+x^{3}+x^{2}+x+1} $$ & (4 Marks) \ \begin{tabular}{ll} c. $$ \lim _{b ightarrow-3} \frac{b+1}{b+3} $$ & (2 Marks) \ d. $$ \lim _{x ightarrow 7} \frac{\sqrt{x+2}-3}{x-7} $$ & (5 Marks) \ \hline\end{tabular} \ \hline\end{tabular} \ \hline\end{tabular}

Question

\begin{tabular}{l} Question $$ \mathbf{5} $$ \ \begin{tabular}{ll} 5.1 Find the limit of each of the functions given below. & $$ \mathbf{2 0} $$ Marks \ a. $$ \lim _{h ightarrow 0} \frac{(h-1)^{3}+1}{h} $$ & (4 Marks) \ b. $$ \lim _{x ightarrow \infty} \frac{4 x^{4}-3 x^{2}+1}{2 x^{4}+x^{3}+x^{2}+x+1} $$ & (4 Marks) \ \begin{tabular}{ll} c. $$ \lim _{b ightarrow-3} \frac{b+1}{b+3} $$ & (2 Marks) \ d. $$ \lim _{x ightarrow 7} \frac{\sqrt{x+2}-3}{x-7} $$ & (5 Marks) \ \hline\end{tabular} \ \hline\end{tabular} \ \hline\end{tabular}

UpStudy AI · Accepted Answer

Let's solve each limit step by step.

### a. $$ \lim _{h \rightarrow 0} \frac{(h-1)^{3}+1}{h} $$

First, we will simplify the expression $$ (h-1)^{3}+1 $$ and then evaluate the limit as $$ h $$ approaches 0.

1. Expand $$ (h-1)^{3} $$:
   $$
   (h-1)^{3} = h^{3} - 3h^{2} + 3h - 1
   $$
   Therefore,
   $$
   (h-1)^{3} + 1 = h^{3} - 3h^{2} + 3h - 1 + 1 = h^{3} - 3h^{2} + 3h
   $$

2. Substitute back into the limit:
   $$
   \lim_{h \rightarrow 0} \frac{h^{3} - 3h^{2} + 3h}{h}
   $$

3. Simplify:
   $$
   = \lim_{h \rightarrow 0} (h^{2} - 3h + 3)
   $$

4. Evaluate the limit:
   $$
   = 0^{2} - 3(0) + 3 = 3
   $$

### b. $$ \lim _{x \rightarrow \infty} \frac{4 x^{4}-3 x^{2}+1}{2 x^{4}+x^{3}+x^{2}+x+1} $$

To find this limit, we will divide the numerator and the denominator by $$ x^{4} $$:

1. Rewrite the limit:
   $$
   \lim_{x \rightarrow \infty} \frac{4 - \frac{3}{x^{2}} + \frac{1}{x^{4}}}{2 + \frac{1}{x} + \frac{1}{x^{2}} + \frac{1}{x^{3}} + \frac{1}{x^{4}}}
   $$

2. As $$ x $$ approaches infinity, the terms with $$ \frac{1}{x} $$ and higher powers will approach 0:
   $$
   = \frac{4 - 0 + 0}{2 + 0 + 0 + 0 + 0} = \frac{4}{2} = 2
   $$

### c. $$ \lim _{b \rightarrow -3} \frac{b+1}{b+3} $$

1. Substitute $$ b = -3 $$:
   $$
   = \frac{-3 + 1}{-3 + 3} = \frac{-2}{0}
   $$
   This indicates that the limit does not exist (it approaches negative or positive infinity).

### d. $$ \lim _{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7} $$

This limit is in the indeterminate form $$ \frac{0}{0} $$. We can apply L'Hôpital's Rule or simplify it.

1. Substitute $$ x = 7 $$:
   $$
   = \frac{\sqrt{7+2}-3}{7-7} = \frac{3-3}{0} = \frac{0}{0}
   $$

2. Apply L'Hôpital's Rule:
   - Differentiate the numerator: $$ \frac{d}{dx}(\sqrt{x+2}) = \frac{1}{2\sqrt{x+2}} $$
   - Differentiate the denominator: $$ \frac{d}{dx}(x-7) = 1 $$

3. Now apply the limit:
   $$
   = \lim_{x \rightarrow 7} \frac{\frac{1}{2\sqrt{x+2}}}{1} = \frac{1}{2\sqrt{7+2}} = \frac{1}{2\sqrt{9}} = \frac{1}{6}
   $$

### Summary of Results:
- a. $$ 3 $$
- b. $$ 2 $$
- c. Does not exist
- d. $$ \frac{1}{6} $$
The limits are: - a. 3 - b. 2 - c. Does not exist - d. 1/6

Question

5.1 Find the limit of each of the functions given below. Marks

a. (4 Marks)

b. (4 Marks)

c. (2 Marks)

d. (5 Marks)

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Solución

a.

b.

c.

d.

Summary of Results:

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Question 5.1 Find the limit of each of the functions given below. Marks a. (4 Marks) b. (4 Marks) c. (2 Marks) d. (5 Marks)