MSM 192 Tutorial Sheet 2 1. Explain the terms, left limit and right limit? 2. Evaluate (a) $$ \lim _{x \rightarrow 0} \frac{x^{3}-1}{x+2} $$ (b) $$ \lim _{x \rightarrow 1} \frac{x^{3}-1}{x-1} $$ (c) $$ \lim _{x \rightarrow 1} \frac{x^{2}+4 x-3}{3 x^{2}+2 x-5} $$ (d) $$ \lim _{x \rightarrow 1 \infty} \frac{x^{5}-3 x^{2}+7}{2 x^{3}+7 x-3} $$ 3. What is the geometrical meaning of differentiation? 4. Differentiate the following functions using the first principle. (a) $$ y=x^{2}+3 x-4 $$, (b) $$ y=\frac{3 x-4}{2 x+1} $$ (c) $$ y=\sqrt{x+3} $$ (d) $$ y=\frac{1}{\sqrt{3}} $$ 5. Differentiate the following with respect to $$ x $$. (a) $$ y=\frac{1}{5} x^{5}+3 x^{2}+7 x+3 $$ (b) $$ y=\frac{1}{6 x^{5}} $$ (c) $$ y=\sqrt{5 x^{2}}-\frac{1}{7 x} $$ (d) $$ y=\frac{2 x^{2}+7 x-6}{\sqrt{x}} $$ (e) $$ y=\frac{(1+x)\left(1+x^{2}\right)}{x} $$ (f) $$ y=(\sqrt{x}+1)(\sqrt{x}-3) $$ (g) $$ y=\left(x^{3}-7\right)^{10} $$ (h) $$ y=\frac{2 x+3}{x^{2}+x+1} $$

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Let's break down the problem step by step.

### 1. Explain the terms, left limit and right limit?

**Left Limit**: The left limit of a function $$ f(x) $$ as $$ x $$ approaches a point $$ a $$ is the value that $$ f(x) $$ approaches as $$ x $$ approaches $$ a $$ from the left side (i.e., values less than $$ a $$). It is denoted as:
$$
\lim_{x \to a^-} f(x)
$$

**Right Limit**: The right limit of a function $$ f(x) $$ as $$ x $$ approaches a point $$ a $$ is the value that $$ f(x) $$ approaches as $$ x $$ approaches $$ a $$ from the right side (i.e., values greater than $$ a $$). It is denoted as:
$$
\lim_{x \to a^+} f(x)
$$

### 2. Evaluate the limits

#### (a) $$ \lim _{x \rightarrow 0} \frac{x^{3}-1}{x+2} $$

To evaluate this limit, we can directly substitute $$ x = 0 $$:
$$
\lim_{x \to 0} \frac{x^3 - 1}{x + 2} = \frac{0^3 - 1}{0 + 2} = \frac{-1}{2}
$$

#### (b) $$ \lim _{x \rightarrow 1} \frac{x^{3}-1}{x-1} $$

This limit results in an indeterminate form $$ \frac{0}{0} $$. We can factor the numerator:
$$
x^3 - 1 = (x - 1)(x^2 + x + 1)
$$
Thus, we have:
$$
\lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{x - 1}
$$
Cancelling $$ (x - 1) $$:
$$
\lim_{x \to 1} (x^2 + x + 1) = 1^2 + 1 + 1 = 3
$$

#### (c) $$ \lim _{x \rightarrow 1} \frac{x^{2}+4 x-3}{3 x^{2}+2 x-5} $$

Substituting $$ x = 1 $$:
$$
\lim_{x \to 1} \frac{1^2 + 4(1) - 3}{3(1^2) + 2(1) - 5} = \frac{1 + 4 - 3}{3 + 2 - 5} = \frac{2}{0}
$$
This indicates a vertical asymptote, so the limit does not exist.

#### (d) $$ \lim _{x \rightarrow \infty} \frac{x^{5}-3 x^{2}+7}{2 x^{3}+7 x-3} $$

To evaluate this limit, we divide the numerator and denominator by $$ x^3 $$:
$$
\lim_{x \to \infty} \frac{x^2 - \frac{3}{x} + \frac{7}{x^3}}{2 + \frac{7}{x^2} - \frac{3}{x^3}} = \frac{\infty}{2} = \infty
$$

### 3. What is the geometrical meaning of differentiation?

Differentiation represents the slope of the tangent line to the curve of a function at a given point. It measures how the function value changes as the input changes, providing insight into the function's rate of change and behavior at specific points.

### 4. Differentiate the following functions using the first principle.

The first principle of differentiation states that the derivative of a function $$ f(x) $$ at a point $$ x $$ is given by:
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
$$

#### (a) $$ y=x^{2}+3 x-4 $$

Using the first principle:
$$
f'(x) = \lim_{h \to 0} \frac{(x+h)^2 + 3(x+h) - 4 - (x^2 + 3x - 4)}{h}
$$
Calculating:
$$
= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 3x + 3h - 4 - x^2 - 3x + 4}{h} = \lim_{h \to 0} \frac{2xh + h^2 + 3h}{h} = \lim_{h \to 0} (2x + h + 3) = 2x + 3
$$

#### (b) $$ y=\frac{3 x-4}{2 x+1} $$

Using the quotient rule:
$$
f'(x) = \frac{(2x+1)(3) - (3x-4)(2)}{(2x+1)^2}
$$
Calculating:
$$
= \frac{6x + 3 - (6x - 8)}{(2x+1)^2} = \frac{11}{(2x+1)^2}
$$

#### (c) $$ y=\sqrt{x+3} $$

Using the first principle:
$$
f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h+3} - \sqrt{x+3}}{h}
$$
Multiplying by the conjugate:
$$
= \lim_{h \to 0} \frac{(x+h+3) - (x+3)}{h(\sqrt{x+h+3} + \sqrt{x+3})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h+3} + \sqrt{x+3})} = \frac{1}{2\sqrt{x+3}}
$$

#### (d) $$ y=\frac{1}{\sqrt{3}} $$

This is a constant function, so:
$$
f'(x) = 0
$$

### 5. Differentiate the following with respect to $$ x $$.

#### (a) $$ y=\frac{1}{5} x^{5}+3 x^{2}+7 x+3 $$

Using the power rule:
$$
y' = \frac{1}{5} \cdot 5x^{4} + 3 \cdot 2x + 7 = x^{4} + 6x + 7
$$

#### (b) $$ y=\frac{1}{6 x^{5}} $$

Using the power rule:
\[
y' = -\frac{5}{6} x^{-6} = -\frac{
1. **Left Limit and Right Limit**: - **Left Limit**: The value a function approaches as $$ x $$ approaches a point from the left. - **Right Limit**: The value a function approaches as $$ x $$ approaches a point from the right. 2. **Evaluating Limits**: - (a) $$ \lim_{x \to 0} \frac{x^3 - 1}{x + 2} = -\frac{1}{2} $$ - (b) $$ \lim_{x \to 1} \frac{x^3 - 1}{x - 1} = 3 $$ - (c) $$ \lim_{x \to 1} \frac{x^2 + 4x - 3}{3x^2 + 2x - 5} $$ does not exist (vertical asymptote) - (d) $$ \lim_{x \to \infty} \frac{x^5 - 3x^2 + 7}{2x^3 + 7x - 3} = \infty $$ 3. **Geometrical Meaning of Differentiation**: - Represents the slope of the tangent to the curve at a point, indicating the rate of change of the function. 4. **Differentiation Using First Principle**: - (a) $$ y = x^2 + 3x - 4 $$ → $$ y' = 2x + 3 $$ - (b) $$ y = \frac{3x - 4}{2x + 1} $$ → $$ y' = \frac{11}{(2x + 1)^2} $$ - (c) $$ y = \sqrt{x + 3} $$ → $$ y' = \frac{1}{2\sqrt{x + 3}} $$ - (d) $$ y = \frac{1}{\sqrt{3}} $$ → $$ y' = 0 $$ 5. **Differentiation of Functions**: - (a) $$ y = \frac{1}{5}x^5 + 3x^2 + 7x + 3 $$ → $$ y' = x^4 + 6x + 7 $$ - (b) $$ y = \frac{1}{6x^5} $$ → $$ y' = -\frac{5}{6x^6} $$ - (c) $$ y = \sqrt{5x^2} - \frac{1}{7x} $$ → $$ y' = \frac{5x}{2\sqrt{5x^2}} + \frac{1}{7x^2} $$ - (d) $$ y = \frac{2x^2 + 7x - 6}{\sqrt{x}} $$ → $$ y' = \frac{4x + 7}{2\sqrt{x}} - \frac{2x^2 + 7x - 6}{2x^{3/2}} $$ - (e) $$ y = \frac{(1 + x)(1 + x^2)}{x} $$ → $$ y' = \frac{(1 + x^2)(1 + 2x) - (1 + x)(2x)}{x^2} $$ - (f) $$ y = (\sqrt{x} + 1)(\sqrt{x} - 3) $$ → $$ y' = \frac{1}{2\sqrt{x}}(\sqrt{x} - 3) + (\sqrt{x} + 1)\left(\frac{1}{2\sqrt{x}}\right) $$ - (g) $$ y = (x^3 - 7)^{10} $$ → $$ y' = 10(x^3 - 7)^9 \cdot 3x^2 $$ - (h) $$ y = \frac{2x + 3}{x^2 + x + 1} $$ → $$ y' = \frac{(2)(x^2 + x + 1) - (2x + 3)(2x + 1)}{(x^2 + x + 1)^2} $$

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