Question

$\color{green} \displaystyle \rm\lim_{x \to1} \frac{ {x}^{100} - 2x + 1}{ {x}^{50} - 2x + 1}$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle \rm\lim_{x \to1} \: \frac{ {x}^{100} - 2x + 1}{ {x}^{50} - 2x + 1}$

If we substitute directly x = 1, we get

$\rm \: = \: \dfrac{1 - 2 + 1}{1 - 2 + 1}$

$\rm \: = \: \dfrac{2 - 2}{2 - 2}$

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

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle \rm\lim_{x \to1} \: \frac{ {x}^{100} - 2x + 1}{ {x}^{50} - 2x + 1}$

can be further rewritten as

$\rm \: = \: \displaystyle \rm\lim_{x \to1} \: \frac{ {x}^{100} - 1 + 1 - 2x + 1}{ {x}^{50} - 1 + 1 - 2x + 1}$

$\rm \: = \: \displaystyle \rm\lim_{x \to1} \: \frac{ {x}^{100} - 1 - 2x + 2}{ {x}^{50} - 1 - 2x + 2}$

$\rm \: = \: \displaystyle \rm\lim_{x \to1} \: \frac{( {x}^{100} - 1) - 2(x - 1)}{ ({x}^{50} - 1) - 2(x - 1)}$

Divide numerator and denominator by x - 1, we get

$\rm \: = \: \displaystyle \rm\lim_{x \to1} \: \dfrac{\dfrac{ {x}^{100} - 1}{x - 1} - 2}{\dfrac{ {x}^{50} - 1}{x - 1} - 2}$

We know,

$\boxed{ \rm{ \:\displaystyle \rm\lim_{x \: \to \: a} \: \frac{ {x}^{n} - {a}^{n}}{x - a} \: = \: {na}^{n - 1} \: }}$

$\rm \: = \: \dfrac{ {100(1)}^{100 - 1} - 2}{ {50(1)}^{50 - 1} - 2}$

$\rm \: = \: \dfrac{ {100(1)}^{99} - 2}{ {50(1)}^{49} - 2}$

$\rm \: = \: \dfrac{100 - 2}{ 50 - 2}$

$\rm \: = \: \dfrac{98}{ 48}$

$\rm \: = \: \dfrac{49}{24}$

Hence,

$\rm\implies \:\displaystyle \rm\lim_{x \to1} \frac{ {x}^{100} - 2x + 1}{ {x}^{50} - 2x + 1} = \: \dfrac{49}{24}$

$\rule{190pt}{2pt}$

Alternative Method :-

Given expression is

$\rm \: \displaystyle \rm\lim_{x \to1} \frac{ {x}^{100} - 2x + 1}{ {x}^{50} - 2x + 1}$

On substituting directly the value of x = 1, we get

$\rm \: = \: \dfrac{1 - 2 + 1}{1 - 2 + 1}$

$\rm \: = \: \dfrac{2 - 2}{2 - 2}$

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

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle \rm\lim_{x \to1} \frac{ {x}^{100} - 2x + 1}{ {x}^{50} - 2x + 1}$

On applying L Hospital Rule, we get

$\rm \: = \: \displaystyle \rm\lim_{x \to1} \frac{\dfrac{d}{dx}( {x}^{100} - 2x + 1)}{ \dfrac{d}{dx}({x}^{50} - 2x + 1)}$

$\rm \: = \: \displaystyle \rm\lim_{x \to1} \: \frac{100 {x}^{99} - 2 + 0 }{ {50x}^{49} - 2 + 0}$

$\rm \: = \: \displaystyle \rm\lim_{x \to1} \: \frac{100 {x}^{99} - 2}{ {50x}^{49} - 2}$

$\rm \: = \: \dfrac{100 - 2}{ 50 - 2}$

$\rm \: = \: \dfrac{98}{ 48}$

$\rm \: = \: \dfrac{49}{24}$

Hence,

$\color{green}\rm\implies \:\displaystyle \rm\lim_{x \to1} \frac{ {x}^{100} - 2x + 1}{ {x}^{50} - 2x + 1} = \: \dfrac{49}{24}$

Remark :-

$\boxed{ \rm{ \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}} ,\: \: \boxed{ \rm{ \:\dfrac{d}{dx}k = 0}} ,\: \: \boxed{ \rm{ \:\dfrac{d}{dx}x = 1}}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$