Question

(i) lim x2 – 3x + 2
x + 1x2 - 4x + 3​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{\red{Solution}}}}}}$

$\Large{\underline{\mathfrak{\bf{\pink{Given}}}}}$

$:\mapsto\sf{\orange{\displaystyle\lim_{x \to 1} \left(\dfrac{x^2-3x+2}{x^3-4x+3}\right)}}$

First , factories of numerators and denominator

$:\mapsto\sf{\displaystyle \lim_{x \to 1} \left(\dfrac{x^2-2x-x+2}{x^3-x^2+x^2-x-3x+3}\right)} \\ \\ \\ :\mapsto\sf{\displaystyle \lim_{x \to 1} \left(\dfrac{x(x-2)-1(x-2)}{x^2(x-1)+x(x-1)-3(x-1)}\right)} \\ \\ \\ :\mapsto\sf{\displaystyle \lim_{x \to 1} \left(\dfrac{(x-2)\cancel{(x-1)}}{\cancel{(x-1)}(x^2+x-3)}\right)} \\ \\ \\ :\mapsto\sf{\:\displaystyle \lim_{x \to 1} \left(\dfrac{x-2}{x^2+x-3}\right)} \\ \\ \\ \small\sf{\green{\:\:\:\:take\:limit\:at\:x\:to\:1}} \\ \\ \\ :\mapsto\sf{\:\dfrac{(1-2)}{(1^2+1-3)}} \\ \\ \\ :\mapsto\sf{\:\dfrac{-1}{2-3}} \\ \\ \\ :\mapsto\sf{\:\dfrac{-1}{-1}} \\ \\ \\ :\mapsto\sf{\orange{\:1\:\:\:\:\:Ans.}}$

Answer 2

AnswEr :

Given Expression,

$\sf lim \: \: \: \: \:\dfrac{ {x}^{2} - 3x + 2}{ {x}^{3} - 4x + 3} \\ \sf{x \longrightarrow \: 1}$

At x = 1,the given expression tends to 0/0 form.

Here,

• We can either alter the equation accordingly

• We can apply the L'Hospital's Rule

L'HOSPITAL'S RULE

• The numerator and denominator of the given expression are differentiated simultaneously and equated to the precursor equations

To avoid confusion,I will be doing the differentiation part individually.

The Numerator

$\boxed{\begin{minipage}{9 cm} \sf \: \dfrac{dy}{dx} = \dfrac{d( {x}^{2} - 3x + 2)}{dx} \\ \\ \longrightarrow \: \sf \: \dfrac{dy}{dx} = \dfrac{ {dx}^{2} }{dx} - 3 \times \dfrac{dx}{dx} + \dfrac{d(2)}{dx} \\ \\ \longrightarrow \: \sf \: \dfrac{dy}{dx} = 2x - 3 - - - - - (1) \end{minipage}}$

The Denominator

$\boxed{\begin{minipage}{7 cm} \sf \: \dfrac{dy}{dx} = \dfrac{d(x {}^{3} - 4x + 3) }{dx} \\ \\ \longrightarrow \: \sf \: \dfrac{dy}{dx} = {3x}^{2} - 4 - - - - - - - (2) \end{minipage}}$

The expression would be re written as :

$\sf \: lim \: \: \: \: \: \dfrac{2x - 3}{3 {x}^{2} - 4 } \\ \sf{ x \longrightarrow1}$

Putting x = 1,

$\longrightarrow \: \sf \: \dfrac{2(1) - 3}{3(1) {}^{2} - 4} \\ \\ \longrightarrow \: \sf \: 1$

At x = 1,the limit of above expression is 1