Question

question 3
3.1 determine
3.1.1 $\lim_{x \to 0}\frac{x^{2}-3x + 4}{x^{2}+2}$ (2)
3.1.2 $\lim_{x \to 1}\frac{e^{x}}{x^{3}-1}$ (3)
3.2 differentiate the following with regard to $x$:
3.2.1 $y = x\cos5x$ (2)
3.2.2 $y=\sqrt{4x - x^{2}}$ (3)
3.2.3 $y = \ln(\sin e^{3x})$ (3)

Explanation:

Step1: Evaluate limit 3.1.1

Substitute $x = 0$ into $\frac{x^{2}-3x + 4}{x^{2}+2}$.
$\lim_{x
ightarrow0}\frac{x^{2}-3x + 4}{x^{2}+2}=\frac{0^{2}-3\times0 + 4}{0^{2}+2}=\frac{4}{2}=2$

Step2: Evaluate limit 3.1.2

Factor the denominator $x^{3}-1=(x - 1)(x^{2}+x + 1)$. Then $\lim_{x
ightarrow1}\frac{\ln x}{x^{3}-1}=\lim_{x
ightarrow1}\frac{\ln x}{(x - 1)(x^{2}+x + 1)}$. Use L - H rule since it is in $\frac{0}{0}$ form. The derivative of $\ln x$ is $\frac{1}{x}$ and the derivative of $x^{3}-1$ is $3x^{2}$. So $\lim_{x
ightarrow1}\frac{\frac{1}{x}}{3x^{2}}=\lim_{x
ightarrow1}\frac{1}{3x^{3}}=\frac{1}{3}$

Step3: Differentiate 3.2.1

Use the product rule $(uv)^\prime=u^\prime v+uv^\prime$, where $u = x$ and $v=\cos5x$. $u^\prime=1$ and $v^\prime=- 5\sin5x$. Then $y^\prime=\cos5x-5x\sin5x$

Step4: Differentiate 3.2.2

First rewrite $y=(4x - x^{2})^{\frac{1}{2}}$. Use the chain - rule $y^\prime=\frac{1}{2}(4x - x^{2})^{-\frac{1}{2}}\times(4 - 2x)=\frac{2 - x}{\sqrt{4x - x^{2}}}$

Step5: Differentiate 3.2.3

Use the chain - rule. Let $u=\sin e^{3x}$. Then $y=\ln u$, $y^\prime=\frac{1}{u}\times u^\prime$. $u^\prime=\cos e^{3x}\times3e^{3x}$. So $y^\prime=\frac{3e^{3x}\cos e^{3x}}{\sin e^{3x}}=3e^{3x}\cot e^{3x}$

Answer:

3.1.1: $2$
3.1.2: $\frac{1}{3}$
3.2.1: $\cos5x-5x\sin5x$
3.2.2: $\frac{2 - x}{\sqrt{4x - x^{2}}}$
3.2.3: $3e^{3x}\cot e^{3x}$