Hard to make out what you mean without brackets for the denominators
the way you typed it....
1/t√1 is just 1/t, remember terms are separated by addition or subtraction
so you simply have
1/t + t - 1/t
= t
I am sure that is not you meant.
(1/t√1+t - 1 /t) lim
t➡️0
So far I factored out 1/t√1+t - 1/t x √1+t/√1+t
I am having trouble going further
5 answers
No its written as 1 in the numerator then in the denominator it is written as t√1+t this is then subtracted by 1/t lim is t----> 0
How would you factor this?
How would you factor this?
Still not clear enough
if it would be t√1 + t
why would they put in that √1 ?
it would simply become t(1) + t
or 2t
is it
t√(1+t) ?
there is a monumental difference
should I guess at
Limit (1/( t√(1+t) - 1/t ) , as x ---> 0 ???
if it would be t√1 + t
why would they put in that √1 ?
it would simply become t(1) + t
or 2t
is it
t√(1+t) ?
there is a monumental difference
should I guess at
Limit (1/( t√(1+t) - 1/t ) , as x ---> 0 ???
yes that is what it is!! sorry for the confusion
OK THEN
let's add up 1/( t√(1+t) - 1/t
= (1 - √(1+t)/(t√(1+t)
= (1 - √(1+t)/(t√(1+t) * (1 + √(1+t) / (1 + √(1+t)
= (1 - 1 - t)/(t√(1+t)((1 + √(1+t) )
= -t/(t√(1+t)((1 + √(1+t) )
= -1/(√(1+t)((1 + √(1+t) )
so lim -1/(√(1+t)((1 + √(1+t) ) as t ---> 0
= -1/(√1(1 + √1)
= -1/2
let's add up 1/( t√(1+t) - 1/t
= (1 - √(1+t)/(t√(1+t)
= (1 - √(1+t)/(t√(1+t) * (1 + √(1+t) / (1 + √(1+t)
= (1 - 1 - t)/(t√(1+t)((1 + √(1+t) )
= -t/(t√(1+t)((1 + √(1+t) )
= -1/(√(1+t)((1 + √(1+t) )
so lim -1/(√(1+t)((1 + √(1+t) ) as t ---> 0
= -1/(√1(1 + √1)
= -1/2
2 answers