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) $$ y=\frac{1}{\sqrt{3}} $$

Question

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1. The left limit at a point is the value that the function approaches as the variable approaches the point from values less than the point. In contrast, the right limit is the value that the function approaches as the variable approaches the point from values greater than the point.

2.  
**(a)** Evaluate  
$$
\lim_{x \rightarrow 0} \frac{x^3-1}{x+2}
$$  
Since the function is continuous at $$x=0$$ (the denominator $$0+2=2\neq 0$$), substitute $$x=0$$:
$$
\frac{0^3-1}{0+2}=\frac{-1}{2}=-\frac{1}{2}.
$$

**(b)** Evaluate  
$$
\lim_{x \rightarrow 1} \frac{x^3-1}{x-1}.
$$  
Factor the numerator using the difference of cubes formula:
$$
x^3-1=(x-1)(x^2+x+1).
$$  
Cancel the common factor (valid for $$x\neq 1$$):
$$
\lim_{x \rightarrow 1}(x^2+x+1)=1^2+1+1=3.
$$

**(c)** Evaluate  
$$
\lim_{x \rightarrow 1} \frac{x^2+4x-3}{3x^2+2x-5}.
$$  
When $$x=1$$:
$$
\text{Numerator: }1^2+4(1)-3=1+4-3=2, \quad \text{Denominator: }3(1)^2+2(1)-5=3+2-5=0.
$$  
Since the denominator tends to 0 while the numerator is a nonzero constant (2), the fraction does not approach a finite number. To check the behavior, factor the denominator. Find its roots by solving
$$
3x^2+2x-5=0.
$$
The discriminant is
$$
\Delta=2^2-4(3)(-5)=4+60=64,
$$
so the roots are
$$
x=\frac{-2\pm 8}{6}.
$$
One root is
$$
x=\frac{6}{6}=1.
$$
Hence, factoring we can write
$$
3x^2+2x-5=3\left(x-1\right)\left(x+\frac{5}{3}\right).
$$
Near $$x=1$$, the sign of the denominator depends on $$(x-1)$$.  
- If $$x\rightarrow 1^+$$, then $$(x-1)>0$$ and the expression approaches $$+\infty$$.  
- If $$x\rightarrow 1^-$$, then $$(x-1)<0$$ and the expression approaches $$-\infty$$.

Thus, the two-sided limit does not exist since the one-sided limits differ.

**(d)** Evaluate  
$$
\lim_{x \rightarrow \infty} \frac{x^5-3x^2+7}{2x^5+7x-3}.
$$  
For large $$x$$ the highest degree terms dominate. Divide numerator and denominator by $$x^5$$:
$$
\lim_{x \rightarrow \infty} \frac{1-\frac{3}{x^3}+\frac{7}{x^5}}{2+\frac{7}{x^4}-\frac{3}{x^5}}=\frac{1}{2}.
$$

3. The geometrical meaning of differentiation is that the derivative at a point represents the slope of the tangent line to the function's curve at that point. In other words, it gives the instantaneous rate of change of the function with respect to its variable.

4. Differentiate using the first principle, where the derivative is defined as
$$
f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.
$$

**(a)** For $$ y=x^2+3x-4 $$:
$$
f(x)=x^2+3x-4.
$$
Calculate $$ f(x+h) $$:
$$
f(x+h)=(x+h)^2+3(x+h)-4=x^2+2xh+h^2+3x+3h-4.
$$
Find the difference:
$$
f(x+h)-f(x)=(x^2+2xh+h^2+3x+3h-4)-(x^2+3x-4)=2xh+h^2+3h.
$$
Divide by $$h$$:
$$
\frac{f(x+h)-f(x)}{h}=2x+h+3.
$$
Take the limit as $$ h\rightarrow 0 $$:
$$
f'(x)=2x+3.
$$

**(b)** For $$ y=\frac{3x-4}{2x+1} $$:
$$
f(x)=\frac{3x-4}{2x+1}.
$$
Using the first principle,
$$
f'(x)=\lim_{h\rightarrow 0}\frac{\frac{3(x+h)-4}{2(x+h)+1}-\frac{3x-4}{2x+1}}{h}.
$$
To combine the fractions, write
$$
f(x+h)-f(x)=\frac{[3(x+h)-4](2x+1)- (3x-4)[2(x+h)+1]}{[2(x+h)+1](2x+1)}.
$$
Expand the numerator:
$$
\begin{aligned}
N(h) &= [3x+3h-4](2x+1)- (3x-4)[2x+2h+1] \
&= (6x^2+3x+6xh+3h-8x-4) - (6x^2+6xh+3x-8x-8h-4).
\end{aligned}
$$
Carefully grouping like terms shows that most terms cancel. One obtains after cancellation:
$$
N(h)=11h.
$$
Thus,
$$
\frac{f(x+h)-f(x)}{h}=\frac{11h}{h\,[2(x+h)+1](2x+1)}=\frac{11}{[2(x+h)+1](2x+1)}.
$$
Taking the limit as $$h\rightarrow 0$$:
$$
f'(x)=\frac{11}{(2x+1)^2}.
$$

**(c)** For $$ y=\sqrt{x+3} $$:
$$
f(x)=\sqrt{x+3}.
$$
Then,
$$
f(x+h)=\sqrt{x+h+3}.
$$
Thus,
$$
\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+h+3}-\sqrt{x+3}}{h}.
$$
Multiply numerator and denominator by the conjugate:
$$
\frac{\sqrt{x+h+3}-\sqrt{x+3}}{h}\cdot\frac{\sqrt{x+h+3}+\sqrt{x+3}}{\sqrt{x+h+3}+\sqrt{x+3}}=\frac{(x+h+3)-(x+3)}{h\,[\sqrt{x+h+3}+\sqrt{x+3}]}=\frac{h}{h\,[\sqrt{x+h+3}+\sqrt{x+3}]}.
$$
Cancel $$h$$:
$$
\frac{1}{\sqrt{x+h+3}+\sqrt{x+3}}.
$$
Taking the limit as $$ h\rightarrow 0 $$:
$$
f'(x)=\frac{1}{2\sqrt{x+3}}.
$$

**(d)** For $$ y=\frac{1}{\sqrt{3}} $$:
Since this is a constant function, the derivative is
$$
f'(x)=0.
$$
1. **Left Limit and Right Limit:** - **Left Limit:** The value a function approaches as the variable approaches a point from values less than that point. - **Right Limit:** The value a function approaches as the variable approaches a point from values greater than that point. 2. **(a)** $$ \lim_{x \rightarrow 0} \frac{x^3 - 1}{x + 2} = -\frac{1}{2} $$ **(b)** $$ \lim_{x \rightarrow 1} \frac{x^3 - 1}{x - 1} = 3 $$ **(c)** $$ \lim_{x \rightarrow 1} \frac{x^2 + 4x - 3}{3x^2 + 2x - 5} \text{ does not exist} $$ **(d)** $$ \lim_{x \rightarrow \infty} \frac{x^5 - 3x^2 + 7}{2x^5 + 7x - 3} = \frac{1}{2} $$ 3. **Geometrical Meaning of Differentiation:** - The derivative at a point represents the slope of the tangent line to the function's curve at that point, indicating the instantaneous rate of change of the function with respect to its variable. 4. **(a)** $$ \frac{dy}{dx} = 2x + 3 $$ **(b)** $$ \frac{dy}{dx} = \frac{11}{(2x + 1)^2} $$ **(c)** $$ \frac{dy}{dx} = \frac{1}{2\sqrt{x + 3}} $$ **(d)** $$ \frac{dy}{dx} = 0 $$

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