Question

lim x tends to 1
pls koi jaldi se bata do​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Evaluate the following

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1}\rm \bigg( \dfrac{ {x}^{2} - 2}{ {x}^{2} - x} + \frac{1}{ {x}^{3} - 3{x}^{2} + 2x} \bigg)$

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1}\rm \bigg( \dfrac{ {x}^{2} - 2}{ {x}^{2} - x} + \frac{1}{ {x}^{3} - 3{x}^{2} + 2x} \bigg)$

If we substitute directly x = 1, we get

$\rm \:  =  \: \dfrac{1 - 2}{1 - 1} - \dfrac{1}{1 - 3 + 2}$

$\rm \:  =  \: \dfrac{ - 1}{0} - \dfrac{1}{0}$

$\rm \:  =  \: - \infty - \infty$

which is indeterminant form

So, Consider

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1}\rm \bigg( \dfrac{ {x}^{2} - 2}{ {x}^{2} - x} + \frac{1}{ {x}^{3} - 3{x}^{2} + 2x} \bigg)$

can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm\bigg( \dfrac{ {x}^{2} - 2 }{x(x - 1)} - \dfrac{1}{x( {x}^{2} - 3x + 2)} \bigg)$

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm\bigg( \dfrac{ {x}^{2} - 2 }{x(x - 1)} - \dfrac{1}{x( {x}^{2} - 2x - x + 2)} \bigg)$

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm\bigg( \dfrac{ {x}^{2} - 2 }{x(x - 1)} - \dfrac{1}{x[x(x - 2) - 1(x - 2)]} \bigg)$

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm\bigg( \dfrac{ {x}^{2} - 2 }{x(x - 1)} - \dfrac{1}{x(x - 2)(x - 1)} \bigg)$

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm\bigg( \dfrac{({x}^{2} - 2)(x - 2) - 1}{x(x - 1)(x - 2)} \bigg)$

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm\bigg( \dfrac{{x}^{3} - {2x}^{2} - 2x + 4 - 1}{x(x - 1)(x - 2)} \bigg)$

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm\bigg( \dfrac{{x}^{3} -{2x}^{2} - 2x + 3}{x(x - 1)(x - 2)} \bigg)$

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm\bigg( \dfrac{(x - 1)( {x}^{2} - x - 3)}{x(x - 1)(x - 2)} \bigg)$

[ Factorize using Long Division See Note Section Below]

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm\bigg( \dfrac{ {x}^{2} - x - 3}{x(x - 2)} \bigg)$

$\rm \:  =  \: \dfrac{1 - 1 - 3}{1(1 - 2)}$

$\rm \:  =  \: \dfrac{ - 3}{ - 1}$

$\rm \:  =  \: 3$

Hence,

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 1}\rm \bigg( \dfrac{ {x}^{2} - 2}{ {x}^{2} - x} + \frac{1}{ {x}^{3} - 3{x}^{2} + 2x} \bigg) = 3}$

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Note :-

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} - x - 3\:\:}}}\\ {\underline{\sf{x - 1}}}& {\sf{\: {x}^{3}  -  {2x}^{2} - 2x + 3\:\:}} \\{\sf{}}& \underline{\sf{ \: \: -  {x}^{3} + \: {x}^{2}   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \:    \: \: -  {x}^{2} - 2x + 3  \:  \:  \:  \:   \:  \:  \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  \:  \:  \:  \:  {x}^{2}  - x  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \:  \:  \:  \:  \:  \:  \:  \:  - 3x + 3  \:\:}} \\{\sf{}}& \underline{\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:3x  - 3\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \: \:  \:  \:  \:  \:  \:  \: \: \: \: \:   \: 0\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}$

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More to Know

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1 \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1 \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1 \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1 \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga \: }}$