# How do I find this limit?  `lim_(t->0)[(1/t)-(1/(tsqrt(1+t)))]` ``

llltkl | College Teacher | (Level 3) Valedictorian

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To evaluate ` ` `lim_(x->0)[1/t-1/(tsqrt(1+t))]`

` ``[1/t-1/(tsqrt(1+t))]`

= `[sqrt(1+t)/(tsqrt(1+t))-1/(tsqrt(1+t))]`

= `[(sqrt(1+t)-1)/(tsqrt(1+t))]`

= `[((sqrt(1+t)-1)(sqrt(1+t)+1))/(tsqrt(1+t)(sqrt(1+t)+1))]`

= `[[(1+t-1))/(tsqrt(1+t)(sqrt(1+t)+1))]`

=`[t/(tsqrt(1+t)(sqrt(1+t)+1))]`` `

=`[1/(sqrt(1+t)(sqrt(1+t)+1))]`

Therefore, `lim_(t->0)[1/t-1/(tsqrt(1+t))]`

= `lim_(t->0)[1/(sqrt(1+t)(sqrt(1+t)+1))]`

=`1/(sqrt(1+0)(sqrt(1+0)+1))`

= `1/(1+1)`

` `= `1/2`

Therefore evaluation of the given limit yields 1/2 as the final solution.

pramodpandey | College Teacher | (Level 3) Valedictorian

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We have

`lim_(t->0)(1/t-1/(tsqrt(1+t)))```

We know `(1+x)^m=1+mx+(m(m-1)x^2)/2+...`

`Thus`

`(1/t-1/(tsqrt(1+t)))=(sqrt(1+t)-1)/(tsqrt(1+t))`

`=((1+t)^(1/2)-1)/(tsqrt(1+t))`

`Apply above binomial theorem`

`=(1+(1/2)t+((1/2)(1/2-1)/2)t^2+...-1}/(tsqrt(1+t))`

`=(t(1/2-1/8t+......))/(tsqrt(1+t))`

`=(1/2-(1/8)t+...)/(sqrt(1+t))`

`Thus`

`lim_{t->0}(1/t-1/(tsqrt(1+t)))=lim_(t->0)(1/2-(1/8)t+...)/(sqrt(1+t))`

`=1/2`

`Ans.`

oldnick | (Level 1) Valedictorian

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`ERRRRATA-CORRRRIGE:`

LAST LINE:

`lim_(t->0) (1-1/sqrt(1+t))/t= lim_(x->0) -(-1/2) xx 1/((1+t)sqrt(1+t))=-(-1/2)=1/2`

oldnick | (Level 1) Valedictorian

Posted on

`lim_(t->0) [1/t-1/(tsqrt(1+t))]=` `lim_(t->0) 1/t xx[1 -1/sqrt(1+t)]=`

`=lim_(t->0) (1-1/sqrt(1+t))/t`   Now  `lim_(t->0) 1-1/sqrt(1+t)=0`

So its a `0/0 `  form: apling De L'Hopistal's rules:

`lim_(t->0) (1-1/sqrt(1+t))/t= lim_(t->0) -1/2xx 1/((1+t)sqrt(1+t))=` `-1/2`