Question

Limit to be solved without using L'Hôpital's rule.
$\lim \limits_{x \to 2} \dfrac{ {x}^{3} - 3 {x}^{2} + 4 }{ {x}^{4} - 8 {x}^{2} + 16}$
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Answer 1

$\rm\large\underline{\text{Explanation}}$

Since the denominator of $\dfrac{x^{3}-3x^{2}+4}{x^{4}-8x^{2}+16}$ becomes zero at $\text{ x=2 , x=-2 ,}$ the value of the function does not exist for $\text{ x=2 , x=-2 .}$ So, substitution does not give the limiting value.

Since rational functions are continuous everywhere except where the denominator becomes zero, we know that,

$\cdots \longrightarrow \displaystyle \lim_{x \to 2-}f(x)=\lim_{x \to 2+}f(x).$

So, the limit exists.

$\boxed{\begin{aligned}\displaystyle\lim_{x \to 2}\dfrac{x^{3}-3x^{2}+4}{x^{4}-8x^{2}+16}&=\lim_{x \to 2}\dfrac{\cancel{(x-2)^{2}}(x+1)}{(x+2)^{2}\cancel{(x-2)^{2}}}\\\\&=\lim_{x \to 2}\dfrac{x+1}{(x+2)^{2}}\ [\because \displaystyle \lim_{x \to a}f(x)g(x)=\lim_{x \to a}f(x) \lim_{x \to a}g(x)]\\\\&=\dfrac{3}{16}\end{aligned}}$

So,

$\cdots \longrightarrow \boxed{\displaystyle\lim_{x \to 2}\dfrac{x^{3}-3x^{2}+4}{x^{4}-8x^{2}+16}=\dfrac{3}{16}}$

$\rm\large\underline{\text{Extra information}}$

Properties of limits

$\boxed{\begin{aligned}&(1)\displaystyle \lim_{x \to a}kf(x)=k \lim_{x \to a}f(x) \\\\&(2)\displaystyle \lim_{x \to a}\{f(x) \pm g(x)\}= \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x)\\\\&(3)\displaystyle \lim_{x \to a}f(x)g(x)=\lim_{x \to a}f(x) \lim_{x \to a} g(x)\\\\&(4)\displaystyle \lim_{x \to a}\dfrac{f(x)}{g(x)}=\frac{\displaystyle \lim_{x \to a}f(x)}{\displaystyle \lim_{x \to a}g(x)}\text{ for \lim_{x \to a}g(x) \neq 0 }\end{aligned}}$

And it holds for $\text{ x \to a+ , x \to a- , x \to \infty , x \to \infty .}$

Answer 2

Question:-

Limit to be solved without using L'Hôpital's rule.

$\sf \large\lim \limits_{x \to 2} \dfrac{ {x}^{3} - 3 {x}^{2} + 4 }{ {x}^{4} - 8 {x}^{2} + 16} .$

Given:-

$\sf \large\lim \limits_{x \to 2} \dfrac{ {x}^{3} - 3 {x}^{2} + 4 }{ {x}^{4} - 8 {x}^{2} + 16} .$

To Find:-

• We have to Find value of the limit.

Solution:-

Let, the value of the limit be z.

$\sf \large\Rightarrow z = \lim \limits_{x \to 2} \dfrac{ {x}^{3} - 3 {x}^{2} + 4 }{ {x}^{4} - 8 {x}^{2} + 16}$

If we put directly x = 2, we get

$\sf \large \frac{0}{0} form.$

So, by L'Hôpital's rule.

$\sf \large \Rightarrow z = \lim \limits_{x \to 2} \: \frac{f'(x)}{g'(x)}$

$\sf \large \Rightarrow z = \lim \limits_{x \to 2} \dfrac{ 3 {x}^{2} - 6x}{ 4x {}^{3} - 16x}$

So, again it occurs

$\sf \large \frac{0}{0}$So Using L'Hôpital's rule again.

$\sf \large \Rightarrow z = \lim \limits_{x \to 2} \: \frac{6x - 6}{12x {}^{2} - 16}$

$\sf \large \Rightarrow z = \frac{6}{32}$

$\sf \large \Rightarrow z = \frac{3}{16}$

Answer:-

$\sf \large \green{Hence, \: the \: value \: of \: limit \: is \: \frac{3}{16}. }$

Hope you have satisfied. ⚘