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

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate the given limit:

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

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

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

Put x = 0, we get:

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

$= \sf \dfrac{3}{2}$

Therefore:

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

$\textsf{\large{\underline{Answer}:}}$

• Result after evaluating the given limit = 3/2.

$\textsf{\large{\underline{Additional Information}:}}$

$\displaystyle\sf 1.\lim_{x\to 0} sin(x)= 0$

$\displaystyle \sf 2.\lim_{x\to 0} cos(x) =1$

$\displaystyle \sf 3.\lim_{x\to 0} \dfrac{sin(x)}{x}= 1$

$\displaystyle \sf 4.\lim_{x\to 0} \dfrac{tan(x)}{x}= 1$

$\displaystyle \sf 5.\lim_{x\to 0} \dfrac{log(1+x)}{x}= 1$

$\displaystyle \sf 6. \lim_{x\to 0}e^{x}= 1$

$\displaystyle \sf 7.\lim_{x\to 0} \dfrac{e^{x}-1}{x}= 1$

$\displaystyle \sf 8.\lim_{x\to 0} \dfrac{a^{x}-1}{x}= ln(a)$

$\displaystyle \sf 9.\lim_{x\to \infty}\bigg(1+\dfrac{1}{x}\bigg)^{x}=e$

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

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