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Question

Find:

(i)limx2[x]

(ii)limx52[x]

(iii)limx1[x]

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Solution

(i)limx2[x]

LHL:

limx2[X]

Let x =2-h,

h=2x

as x20x<2 slightly

2x>0h>0

h0+

limh0+[2h]=1

RHLlimx2+[X]

Let x=2+h,h=x2

ax x2+x>2 slightly

2>0h>0

h0+

limh0+[2+h]

LHLRHL

Thus, limx2[x] does not exist.

(ii)limx52[x]

RHL:limx5+2[x]

Let x=52+h,

h=x52

as x5+2x>52 slightly

x52>0

h>0

h0+

limh0+[52+h]=2

Hence, limx52[x]

Let x=52h,h=52x

as x52x<52 slightly

52x>0

h>0

h0+

limh0+[52h]=2

Hence, limx52[x]=2

Since, limx52[x]=limx5+2[x]

limx52[x]=2

(iii)limx1[x]

LHL:limx1[x]

Let x =1-h,

h=1x

as x1x<1 slightly

1x>0h>0

h0+

limh0+[1h]

limx1[x]=0

RHL:limx1+[x]

Let x =1+h,

h=x1

as x1+x>1 slightly

x1>0h>0

h0+

limx0+[1+4]=1

limx1[x]limx1+[x]

Thus, limx1[x] does not exist.


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