Question

$\huge \bold \red{\sf\displaystyle \lim_{x \to 0} \bigg(\dfrac{x^{2} + 3x }{x^{3} + 2x} \bigg)}$



❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​​​​​​​​​​​​​​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \lim_{x \to 0} \bigg(\dfrac{x^{2} + 3x }{x^{3} + 2x} \bigg)$

$= \displaystyle \lim_{x \to 0} \dfrac{x(x+ 3) }{x( {x}^{2} + 2)}$

$= \displaystyle \lim_{x \to 0} \dfrac{x+ 3}{{x}^{2} + 2}$

$= \dfrac{0+ 3}{0 + 2}$

$= \dfrac{ 3}{ 2}$

Answer 2

$\lim_{x \to 0} \frac{x^{2} + 3 x}{x^{3} + 2 x}$

Factor:

$\color{red}{\lim_{x \to 0} \frac{x^{2} + 3 x}{x^{3} + 2 x}} = \color{red}{\lim_{x \to 0} \frac{x \left(x + 3\right)}{x \left(x^{2} + 2\right)}}$

Cancel the common term:

$\color{red}{\lim_{x \to 0} \frac{x \left(x + 3\right)}{x \left(x^{2} + 2\right)}} = \color{red}{\lim_{x \to 0} \frac{x + 3}{x^{2} + 2}}$

Substitute the variable with the value:

$\color{red}{\lim_{x \to 0} \frac{x + 3}{x^{2} + 2}} = \color{red}{\left(\frac{3}{2}\right)}$

Therefore,

$\lim_{x \to 0} \frac{x^{2} + 3 x}{x^{3} + 2 x} = \frac{3}{2}$

Answer:

$\lim_{x \to 0} \frac{x^{2} + 3 x}{x^{3} + 2 x}=\frac{3}{2}$