Question

lim x-1 logx/x-1
find the correct solution for it​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1}\rm \frac{logx}{x - 1}$

If we substitute directly x = 1, we get

$\rm \:  =  \: \dfrac{log1}{1 - 1}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form

So,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1}\rm \frac{logx}{x - 1}$

To evaluate this limit, we use Method of Substitution

So, Substitute

$\purple{\rm :\longmapsto\:x = 1 + y}$

$\rm\implies \:as \: x \to \: 1 \: \: so \: y \to \: 0$

So, above expression can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{y \to 0}\rm \frac{log(1 + y)}{1 + y - 1}$

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

$\rm \:  =  \: 1$

Hence,

$\\ \purple{\rm\implies \:\boxed{\tt{ \: \: \: \displaystyle\lim_{x \to 1}\rm \frac{logx}{x - 1} = 1 \: \: \: }}}$

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MORE TO KNOW

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

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

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

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

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

Answer 2

Step-by-step explanation:

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