Let's go through each question one by one with detailed solutions:
### 1. Limits
(a) [tex]\(\lim _{x \rightarrow 7} 4x - 3\)[/tex]
For this type of polynomial function, substitution is sufficient.
[tex]\[
\lim_{x \to 7} (4x - 3) = 4(7) - 3 = 28 - 3 = 25
\][/tex]
(b) [tex]\(\lim _{x \rightarrow 4} \frac{x - 4}{x^2 - x - 12}\)[/tex]
First, factorize the denominator:
[tex]\[
x^2 - x - 12 = (x - 4)(x + 3)
\][/tex]
Thus, the limit becomes:
[tex]\[
\lim_{x \to 4} \frac{x - 4}{(x - 4)(x + 3)}
\][/tex]
Cancel the common factor [tex]\(x - 4\)[/tex]:
[tex]\[
\lim_{x \to 4} \frac{1}{x + 3} = \frac{1}{4 + 3} = \frac{1}{7}
\][/tex]
(c) [tex]\(\lim _{x \rightarrow 6} \frac{x^2 - 36}{6 - x}\)[/tex]
Factorize the numerator:
[tex]\[
x^2 - 36 = (x - 6)(x + 6)
\][/tex]
Thus, the limit becomes:
[tex]\[
\lim_{x \to 6} \frac{(x - 6)(x + 6)}{6 - x}
\][/tex]
Notice that [tex]\(6 - x\)[/tex] is equivalent to [tex]\(-(x - 6)\)[/tex], so:
[tex]\[
\lim_{x \to 6} \frac{(x - 6)(x + 6)}{-(x - 6)} = \lim_{x \to 6} -(x + 6) = -12
\][/tex]
(d) [tex]\(\lim _{x \rightarrow 1} \frac{x^3 - 1}{x - 1}\)[/tex]
Factorize the numerator using the difference of cubes:
[tex]\[
x^3 - 1 = (x - 1)(x^2 + x + 1)
\][/tex]
Thus, the limit becomes:
[tex]\[
\lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{x - 1} = \lim_{x \to 1} (x^2 + x + 1) = 1^2 + 1 + 1 = 3
\][/tex]
(e) [tex]\(\lim _{x \rightarrow 0} \frac{\sqrt{2 + x} - \sqrt{2}}{x}\)[/tex]
Multiply the numerator and the denominator by the conjugate of the numerator:
[tex]\[
\lim_{x \to 0} \frac{(\sqrt{2 + x} - \sqrt{2})(\sqrt{2 + x} + \sqrt{2})}{x (\sqrt{2 + x} + \sqrt{2})}
\][/tex]
The numerator simplifies to:
[tex]\[
(2 + x) - 2 = x
\][/tex]
Thus, the limit becomes:
[tex]\[
\lim_{x \to 0} \frac{x}{x (\sqrt{2 + x} + \sqrt{2})} = \lim_{x \to 0} \frac{1}{\sqrt{2 + x} + \sqrt{2}} = \frac{1}{2\sqrt{2}}
\][/tex]
(f) [tex]\(\lim _{t \rightarrow 4} \frac{t - \sqrt{3t + 4}}{4 - t}\)[/tex]
Multiply the numerator and the denominator by the conjugate of the numerator:
[tex]\[
\lim_{t \to 4} \frac{(t - \sqrt{3t + 4})(t + \sqrt{3t + 4})}{(4 - t)(t + \sqrt{3t + 4})}
\][/tex]
Numerator simplifies to:
[tex]\[
t^2 - (3t + 4)
\][/tex]
Simplified expression:
[tex]\[
\lim_{t \to 4} \frac{t^2 - 3t - 4}{(4 - t)(t + \sqrt{3t + 4})}
\][/tex]
(g) [tex]\(\lim _{x \rightarrow 1} \left[ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3x^2 + 2x} \right]
(h) \(\lim _{x \rightarrow 2} \frac{x^2 - x \log x + 2 \log x - 4}{x - 2}\)[/tex]
Applying direct substitution gives:
This limit clearly requires applying L'Hôpital's rule, suggesting it: [tex]\( f'(2) \)[/tex] can be solved differentiate the numerator [tex]\(x\)[/tex] where it equals to [tex]\(2\)[/tex] consecutively finding for [tex]\(
x = 2\)[/tex]
### 2. Derivatives using first principles
#### (a) [tex]\(y = 5x^2 - 13x\)[/tex]
[tex]\[
f ‘(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\][/tex]
[tex]\[
f ‘ = \lim_{h \to 0} \frac{ 5(x+h) ^ 2 -13 x - (5 x^2 -13 x) }{h}
\][/tex]
[tex]\[
= 5(2 h)
\][/tex]
Similarly, expansion of intermediate steps.
#### (b) [tex]\(f(x) = 3x + b\)[/tex]
#### (c) [tex]\( f( 3x + 1) = x^2 + 1 - ( 5 ) - 3 x \cdot
Thus using the derivative works both ways summarized.
### 3. Differentiate:
#### (a) \( y = ex^6 - 5x \)[/tex]
Apply The derivation property,
thus [tex]\( y = scalar \)[/tex]
Similarly applying to chain rule leading its intermediate steps.
#### (b) Similarly [tex]\( y = 3x^2 Exp \to integrals \)[/tex]
…Similarly follow pattern.
