`lim_(t->0) (1/(tsqrt(1+t)) - 1/t)` Evaluate the limit, if it exists.

Textbook Question

Chapter 2, 2.3 - Problem 29 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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kspcr111 | In Training Educator

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lim_(t->0) (1/(tsqrt(1+t)) - 1/t) 

sol:

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

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

now simplify the numerator

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

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

 so,

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

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

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

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

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

as,the denominator is 

`(lim_(t->0)(1+ sqrt(1+t)(sqrt(1+t))))`

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

as t-> 0 we get

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

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

so,

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

so, from (1)

 we get

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

there fore

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

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