\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)

Upstudy AI Solution

Answer

Solution

a.

b.

c.

d.

Summary of Results:

The Deep Dive

Related Questions

Latest Calculus Questions

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)