Question

lim x-->0
${x}^{3} - 4 {x} \div 2 {x}^{2} + 3x$
pls solve it.​

Answer 1

Solution:

Given Limit:

$\displaystyle \rm = \lim_{x \to0} \dfrac{ {x}^{3} - 4x }{2 {x}^{2} + 3x }$

If we substitute x = 0 directly, we get:

$\displaystyle \rm =\dfrac{ {0}^{3} - 0 }{2 \times {0}^{2} + 3 \times 0 }$

$\displaystyle \rm =\dfrac{0}{0}$

So, consider the limit again:

$\displaystyle \rm = \lim_{x \to0} \dfrac{ {x}^{3} - 4x }{2 {x}^{2} + 3x }$

Can be written as:

$\displaystyle \rm = \lim_{x \to0} \dfrac{x( {x}^{2} - 4) }{x(2x + 3) }$

$\displaystyle \rm = \lim_{x \to0} \dfrac{{x}^{2} - 4}{2x + 3}$

Now, substituting x = 0, we will get the value of the limit:

$\displaystyle \rm =\dfrac{{0}^{2} - 4}{2 \times 0+ 3}$

$\displaystyle \rm = \dfrac{ - 4}{3}$

Therefore:

$\displaystyle \rm \longrightarrow \lim_{x \to0} \dfrac{ {x}^{3} - 4x }{2 {x}^{2} + 3x } = \dfrac{ - 4}{3}$

★ Which is our required answer.

To Know More:

Some standard limits.

$\displaystyle\rm 1.\:\: \lim_{x\to0}\sin(x)=0$

$\displaystyle\rm 2.\:\: \lim_{x\to0}\cos(x)=1$

$\displaystyle\rm 3.\:\: \lim_{x\to0}\dfrac{\sin(x)}{x}=1$

$\displaystyle\rm 4.\:\: \lim_{x\to0}\dfrac{\tan(x)}{x}=1$

$\displaystyle\rm 5.\:\: \lim_{x\to0}\dfrac{1-\cos(x)}{x}=0$

$\displaystyle\rm 6.\:\: \lim_{x\to0}\dfrac{\sin^{-1}(x)}{x}=1$

$\displaystyle\rm 7.\:\: \lim_{x\to0}\dfrac{\tan^{-1}(x)}{x}=1$

$\displaystyle\rm 8.\:\: \lim_{x\to0}\dfrac{\log(1+x)}{x}=1$

$\displaystyle\rm 9.\:\: \lim_{x\to0}\dfrac{e^{x}-1}{x}=1$