1.

lim_(xto0) |x|

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Solution :We see that when `xto0, |x|to0`
Let US establish this using `EPSILON-delta` technique. Let `epsilongt0` we seek a `deltagt0` DEPENDING on
`epsilons.t. |x-0|epsilon implies||x|-0|let epsilon`
Now `||x|-o|=||x||=|x|ltdelta`
Choosing `epsilon=delta` we have
`|x|topsoil implies||x|-0|ltepsilon`
`therefore lim_(xto0)|x|=0`


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