Question

question 3
3.1 determine the following limits:
3.1.1 $lim_{x \to 0}\frac{2x}{cos x}$ (2)
3.1.2 $lim_{x \to infty}\frac{x^{2}+3x + 2}{2x^{2}+x + 3}$ (3)

Explanation:

Step1: Evaluate limit for 3.1.1

Use direct - substitution. Since $\cos(0) = 1$ and the function $\frac{2x}{\cos x}$ is continuous at $x = 0$.
$\lim_{x
ightarrow0}\frac{2x}{\cos x}=\frac{2\times0}{\cos(0)}$

Step2: Calculate the value for 3.1.1

$\frac{2\times0}{1}=0$

Step3: Evaluate limit for 3.1.2

Divide both the numerator and denominator by $x^{2}$ as $x
ightarrow\infty$.
$\lim_{x
ightarrow\infty}\frac{x^{2}+3x + 2}{2x^{2}+x + 3}=\lim_{x
ightarrow\infty}\frac{1+\frac{3}{x}+\frac{2}{x^{2}}}{2+\frac{1}{x}+\frac{3}{x^{2}}}$

Step4: Calculate the value for 3.1.2

As $x
ightarrow\infty$, $\frac{3}{x}
ightarrow0$, $\frac{2}{x^{2}}
ightarrow0$, $\frac{1}{x}
ightarrow0$ and $\frac{3}{x^{2}}
ightarrow0$. So $\lim_{x
ightarrow\infty}\frac{1+\frac{3}{x}+\frac{2}{x^{2}}}{2+\frac{1}{x}+\frac{3}{x^{2}}}=\frac{1 + 0+0}{2+0 + 0}=\frac{1}{2}$

Answer:

3.1.1: $0$
3.1.2: $\frac{1}{2}$