Question

EVALUATE



❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​​​​​​​​​​​​​​​​​​​​

Answer 1

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

Given function is

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: g(x) = \begin{cases} &\sf{ {x}^{2} - 3 \: \: when \: x < 1 } \\ &\sf{2x + 4 \: \: when \: x > 1} \end{cases}\end{gathered}\end{gathered}$

We know, Limit exist at x = 1 iff LHL of g(x) at x = 1 and RHL of g(x) at x = 1 is equal.

That means,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1} \bf \: g(x) \: exist \: iff$

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}g(x) \: = \: \displaystyle\lim_{x \to 1^ + }g(x) \: }$

Consider LHL at x = 1

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}g(x)$

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

To evaluate this limit, we use method of Substitution.

So,

$\red{\rm :\longmapsto\:Put \: x = 1-h, \: as \: x \to 1, \: so \: h \to 0}$

So, above can be rewritten as

$\rm \:  =  \:\displaystyle\lim_{h \to \: 0} \: [ {(1 - h)}^{2} - 3]$

$\rm \:  =  \: {(1 - 0)}^{2} - 3$

$\rm \:  =  \: {1}^{2} - 3$

$\rm \:  =  \: 1 - 3$

$\rm \:  =  \: - 2$

$\rm \implies\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 1^-}g(x) = - 2 \: }}$

Now, Consider RHL at x = 1

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^ + }g(x)$

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

To evaluate this limit, we use method of Substitution.

So,

$\red{\rm :\longmapsto\:Put \: x = 1 + h, \: as \: x \to 1, \: so \: h \to 0}$

So, above can be rewritten as

$\rm \:  =  \:\displaystyle\lim_{h \to \: 0} \: [ 2(1 + h) + 4]$

$\rm \:  =  \:2(1 + 0) + 4$

$\rm \:  =  \:2(1) + 4$

$\rm \:  =  \:2 + 4$

$\rm \:  =  \:6$

So,

$\rm \implies\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 1^ + }g(x) = 6 \: }}$

Thus,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}g(x) \: \ne \: \displaystyle\lim_{x \to 1^ + }g(x)$

So,

$\red{\bf\implies \:\boxed{ \tt{ \: \displaystyle\lim_{x \to 1} \bf \: g(x) \: doesnot \: exist}}}$