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) $$ \lim _{x \rightarrow 1 \infty} \frac{x^{5}-3 x^{2}+7}{2 x^{5}+7 x-3} $$ 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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### 1. Explanation of Left Limit and Right Limit

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

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

### 2. Evaluate the Limits

Let's evaluate each limit step by step.

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

To evaluate this limit, we can directly substitute $$ x = 0 $$:

$$
\lim _{x \rightarrow 0} \frac{x^{3}-1}{x+2} = \frac{0^{3}-1}{0+2} = \frac{-1}{2}
$$

#### (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:

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

Cancelling $$ x-1 $$:

$$
\lim _{x \rightarrow 1} (x^{2}+x+1) = 1^{2}+1+1 = 3
$$

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

Substituting $$ x = 1 $$:

$$
\lim _{x \rightarrow 1} \frac{1^{2}+4(1)-3}{3(1^{2})+2(1)-5} = \frac{1+4-3}{3+2-5} = \frac{2}{0}
$$

This indicates a vertical asymptote, so the limit does not exist.

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

To evaluate this limit, we divide the numerator and denominator by $$ x^{5} $$:

$$
\lim _{x \rightarrow \infty} \frac{1 - \frac{3}{x^{3}} + \frac{7}{x^{5}}}{2 + \frac{7}{x^{4}} - \frac{3}{x^{5}}}
$$

As $$ x $$ approaches infinity, the terms with $$ x $$ in the denominator approach 0:

$$
\lim _{x \rightarrow \infty} \frac{1 - 0 + 0}{2 + 0 - 0} = \frac{1}{2}
$$

### 3. 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 Using the First Principle

The first principle of differentiation states that the derivative of a function $$ f(x) $$ at a point $$ a $$ is given by:

$$
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}
$$

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

Let $$ f(x) = x^{2}+3x-4 $$:

$$
f'(x) = \lim_{h \to 0} \frac{(x+h)^{2}+3(x+h)-4 - (x^{2}+3x-4)}{h}
$$

Calculating this gives:

$$
= \lim_{h \to 0} \frac{(x^{2}+2xh+h^{2}+3x+3h-4) - (x^{2}+3x-4)}{h} = \lim_{h \to 0} \frac{2xh+h^{2}+3h}{h} = \lim_{h \to 0} (2x+h+3) = 2x+3
$$

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

Let $$ f(x) = \frac{3x-4}{2x+1} $$:

Using the quotient rule:

$$
f'(x) = \frac{(2x+1)(3) - (3x-4)(2)}{(2x+1)^{2}}
$$

Calculating this gives:

$$
= \frac{6x+3 - (6x-8)}{(2x+1)^{2}} = \frac{11}{(2x+1)^{2}}
$$

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

Let $$ f(x) = \sqrt{x+3} $$:

$$
f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h+3} - \sqrt{x+3}}{h}
$$

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

$$
\frac{dy}{dx} = x^{4}+6x+7
$$

#### (b) $$ y=\frac{1}{6 x
1. **Left Limit and Right Limit**: - **Left Limit**: The value that a function approaches as $$ x $$ approaches a point from the left. - **Right Limit**: The value that a function approaches as $$ 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^5 + 7x - 3} = \frac{1}{2} $$ 3. **Geometrical Meaning of Differentiation**: - Represents the slope of the tangent to the curve at a point, indicating the rate of change of the function at that point. 4. **Differentiation Using the First Principle**: - (a) $$ y = x^2 + 3x - 4 $$ → $$ \frac{dy}{dx} = 2x + 3 $$ - (b) $$ y = \frac{3x - 4}{2x + 1} $$ → $$ \frac{dy}{dx} = \frac{11}{(2x + 1)^2} $$ - (c) $$ y = \sqrt{x + 3} $$ → $$ \frac{dy}{dx} = \frac{1}{2\sqrt{x + 3}} $$ - (d) $$ y = \frac{1}{\sqrt{3}} $$ → $$ \frac{dy}{dx} = 0 $$ 5. **Differentiation of Given Functions**: - (a) $$ y = \frac{1}{5}x^5 + 3x^2 + 7x + 3 $$ → $$ \frac{dy}{dx} = x^4 + 6x + 7 $$ - (b) $$ y = \frac{1}{6x} $$ → $$ \frac{dy}{dx} = -\frac{1}{6x^2} $$ - (c) $$ y = \sqrt{5x^2} - \frac{1}{7x} $$ → $$ \frac{dy}{dx} = \frac{5x}{\sqrt{5x^2}} + \frac{1}{7x^2} $$ - (d) $$ y = \frac{2x^2 + 7x - 6}{\sqrt{x}} $$ → $$ \frac{dy}{dx} = \frac{4x + 7}{2\sqrt{x}} - \frac{2x^2 + 7x - 6}{2x^{3/2}} $$ - (e) $$ y = \frac{(1 + x)(1 + x^2)}{x} $$ → $$ \frac{dy}{dx} = \frac{(1 + x^2)(1 + 2x) - (1 + x)(2x)}{x^2} $$ - (f) $$ y = (\sqrt{x} + 1)(\sqrt{x} - 3) $$ → $$ \frac{dy}{dx} = \frac{1}{2\sqrt{x}}(\sqrt{x} - 3) + (\sqrt{x} + 1)\frac{1}{2\sqrt{x}} $$ - (g) $$ y = (x^3 - 7)^{10} $$ → $$ \frac{dy}{dx} = 10(x^3 - 7)^9 \cdot 3x^2 $$ - (h) $$ y = \frac{2x + 3}{x^2 + x + 1} $$ → $$ \frac{dy}{dx} = \frac{(2)(x^2 + x + 1) - (2x + 3)(2x + 1)}{(x^2 + x + 1)^2} $$

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