√1+t-√1-t/t lim
-----> 0
The answer is one and I am not getting it
I factored it like this √1+t-√1-t/t x √1+t-√1-t
Factoring it out further I got in the numerator
(1-t)(1+t) / t(√1+t+√1-t) how do you cancel further?
I am going to go out on a limb and guess that you meant:
lim ( √(1+t) - √(1-t) )/t , as t --->0
noting to factor here, you have to rationalize the numerator
= lim ( √(1+t) - √(1-t) )/t * (√(1+t) + √(1-t))/(√(1+t) + √(1-t))
= lim ( 1+t - (1 -t) )/(t(√(1+t) + √(1-t))
= lim 2t/(t(√(1+t) + √(1-t))
= lim 2/(√(1+t) + √(1-t)) as t ---> 0
= 2/(√1 + √1)
= 2/2
= 1
btw, did you clean up your previous post ?
yes i will now
To simplify the expression further, we can proceed with the following steps:
1. Apply the distributive property to expand the expression in the numerator:
(1 - t) * (1 + t) = 1 - t^2
2. Rewrite the denominator as a difference of squares:
√(1 + t) * √(1 - t) = √((1 + t) * (1 - t)) = √(1 - t^2)
3. Substitute the simplified numerator and denominator back into the original expression:
(√(1 + t) - √(1 - t)) / t * (√(1 + t) + √(1 - t)) / √(1 - t^2)
4. Cancel out the common factor of (1 - t^2) in both the numerator and denominator:
(√(1 + t) - √(1 - t)) / t * (√(1 + t) + √(1 - t)) / (√(1 - t) * √(1 + t))
5. Simplify the expression further by canceling out the √(1 - t) in both the numerator and denominator:
(√(1 + t) - √(1 - t)) / t * (√(1 + t) + √(1 - t)) / (√(1 + t))
6. Now, substitute t = 0 into the simplified expression:
(√(1 + 0) - √(1 - 0)) / 0 * (√(1 + 0) + √(1 - 0)) / (√(1 + 0))
We can see that we have a 0 in the denominator, which makes the expression undefined. Therefore, the limit of the expression as it approaches 0 does not exist.