Evaluate lim x-->0 sqrt(3+x) - sqrt(3) / x
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lim f(x) = lim [sqrt(3+x) - sqrt(3)]/x when x--> 0
By substitution:
lim f(x) = 0/0
Now let us multiply and divide by (sqrt(3+x) + sqrt(3)
==> lim [(sqrt(3+x)- sqrt(3))*(sqrt(3+x) + sqrt(3)]/x(sqrt(3+x)+sqrt3).
= lim (3+x - 3)/ x(sqrt(3+x) + sqrt(3)
= lim x/x(sqrt(x+3)+sqrt3)
= lim 1/(sqrt(3+x) + sqrt3)
Then,
lim f(x) when x--> 0 = 1/(sqrt3+0) + sqrt3)
= 1/2sqrt3 = sqrt3/6
Then lim f(x) = sqrt3/6
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First, we'll check to see if we have an indetermination case. For this reason, we'll substitute x by 0.
lim [sqrt(3+x) - sqrt(3)] / x = [sqrt(3+0) - sqrt3]/0 = 0/0
"0/0" is an indetermination, so we could use l'Hospital rule.
If lim (f/g) = 0/0, then lim (f/g)= lim (f')/(g')
lim [sqrt(3+x) - sqrt(3)] / x = lim [sqrt(3+x) - sqrt(3)]' / (x)'
[sqrt(3+x) - sqrt(3)]' = (3+x)'/2sqrt(3+x) - 0
where (3+x)' = 0 + 1 = 1
[sqrt(3+x) - sqrt(3)]' = 1/2sqrt(3+x)
x' = 1
lim [sqrt(3+x) - sqrt(3)] / x = lim 1/1*2sqrt(3+x)
Now, we'll substitute x by 0:
lim 1/1*2sqrt(3+x) = 1/2*sqrt(3+0)
lim 1/1*2sqrt(3+x) = 1/2sqrt3
lim 1/1*2sqrt(3+x) = sqrt3/2*3
lim [sqrt(3+x) - sqrt(3)] / x = sqrt3/6
To find the limit of (sqrt(3+x)-sqrt3)/x as x-->0
Solution:
Substitition x= 0 makes the expression 0/0 of indetermination.
(i)Using differentiating technic we can evaluate the limit.
(ii) The limit could be solved by rationalising the numerator (by multiplying numerator (sqrt(3+x) -sqrt) and denominator x by the conjugate surd , i.e sqrt(( 3+x)+sqrt3 ).
Using diferentiating method:
We know that the definition of differention is given by:
f'(u) = Lt {f(u+x) - f(u)}/x as x-->0
Take f(u) = sqrt u
Then {f'(u) at u= 3} = Lt {sqrt(3+x) -sqrt3}/x as x--> 0
But{ f'(u) at u =3} = { (sqrt(u))' at u =3} = {(1/2)u^(1/2 - 1) at 3 =3 }, as (u^n)' = nu^(n-1)
(f'(u) at u = 3} = (1/2)(3^(1/2 - 1) = 1/(2sqrt3).
Therefore Lt (sqrt(3+x)-sqrt3)/x as x--> 0 = (sqrt u)' at u=3 = 1/(2sqrt3) . Ratinalise the denominator.
Lt(sqrt(3+x)-sqrt3)/x = sqr3/(2sqrt3*sqrt3) = (sqrt3)/6.
Second method:
Lt (sqrt(3+x)-sqrt3)/x = Lt (sqrt(3+x)-sqrt3)(sqrt(3+x)+sqrt3)/{xsqrt(3+x)+sqrt3)
Lt (sqrt(3+x)-sqrt3)/x = Lt ((3+x)-3)/{xsqrt(3+x)+sqrt3)
Lt (sqrt(3+x)-sqrt3)/x = Lt x/{xsqrt(3+x)+sqrt3). We reduce numaratot and denominator by x.
Lt (sqrt(3+x)-sqrt3)/x = Lt 1/{sqrt(3+0)+sqrt3)
Lt (sqrt(3+x)-sqrt3)/x = 1/(2sqrt3) = sqrt3/(2sqrt3*sqrt3) = sqrt3/6.
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