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

Textbook Question

Chapter 2, 2.3 - Problem 25 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

2 Answers | Add Yours

kspcr111's profile picture

kspcr111 | In Training Educator

Posted on

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

sol:

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

Taking the numerator we get 

`sqrt(1 + t) - sqrt(1 - t)`

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

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

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

now taking the (1) we get

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

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

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

now as ` t->0` we get

= `(2/(sqrt(1 + 0) + sqrt(1 - 0)))`

=`2/(1+1)`

= `2/2 = 1`

is the soultion

Wiggin42's profile pic

Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian

Posted on

Plugging in t = 0 into the function results in the indeterminate form 0/0. We could multiply by conjugates and simplify the expression. However, since the numerator and denominator are both differentiable functions, we can apply L'Hopital's rule, which will be easier. Take the derivative of the numerator and denominator. 

`lim_(t->0) ( .5(1 + t)^(-1/2) + .5(1 - t)^(-1/2) )/(1)` Plugging in t = 0 shows that the limit is 1. 

Sources:

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question