Question

directions:

• provide a solution for every problem, place your final answer in a box.
• keep your solution readable.
• sharing of materials such as eraser, pens, paper is not allowed.
• do not disturb others, focus on solving on your own.
• cheating in any form is prohibited.

test i. solve the following:

1. find the value of \\(\lim_{x\to0}\frac{\tan^{2}x}{3x})
2. find the value of \\(\lim_{x\to - 2}\frac{3x^{3}-1x^{2}+4x}{9x^{2}+12x + 4})

Explanation:

Step1: Analyze the first limit

For $\lim_{x
ightarrow0}\frac{\tan^{2}x}{3x}$, use the fact that $\tan x\sim x$ as $x
ightarrow0$. So $\tan^{2}x\sim x^{2}$ as $x
ightarrow0$. Then $\lim_{x
ightarrow0}\frac{\tan^{2}x}{3x}=\lim_{x
ightarrow0}\frac{x^{2}}{3x}$.

Step2: Simplify the first - limit expression

$\lim_{x
ightarrow0}\frac{x^{2}}{3x}=\lim_{x
ightarrow0}\frac{x}{3}=0$.

Step3: Analyze the second limit

For $\lim_{x
ightarrow0}\frac{3x - 12x^{2}}{9x^{3}-x^{2}+4x}$, factor out the greatest - common factor from the numerator and the denominator. The numerator $3x - 12x^{2}=x(3 - 12x)$ and the denominator $9x^{3}-x^{2}+4x=x(9x^{2}-x + 4)$.

Step4: Simplify the second - limit expression

Cancel out the common factor $x$ (since $x
eq0$ when taking the limit as $x
ightarrow0$), we get $\lim_{x
ightarrow0}\frac{3 - 12x}{9x^{2}-x + 4}$. Then substitute $x = 0$ into the expression, $\frac{3-12\times0}{9\times0^{2}-0 + 4}=\frac{3}{4}$.

Answer:

1. $0$
2. $\frac{3}{4}$