levaluate lim x tends to 1 x^2- sqrt x / sqrt x-1?

Answer 1

limit (x^2 - root x) / (root x - 1)
x--->1

= limit root x ( x^3/2 -1) / (root x - 1)
x--->1

= limit root x X limit (x^3/2 - 1) / (root x - 1)
x--->1 x--->1

= 1^1/2 X 3/2 X 1^(3/2 - 1)
[since limit (x^n -a^n) / (x-a) = n*a^(n-1) ]
x--->a

=1 X 3/2 X 1^1/2

=1 X 3/2 X 1

=3/2

Answer 2

Assuming you want to solve

lim [ (x^2 - sqrt(x)) / sqrt(x - 1) ]
x -> 1

One way to solve this without using L'Hospital's rule is to multiply top and bottom by the conjugate of the top. That is, multiply top and bottom by x^2 + sqrt(x). This will give us a difference of squares on the top.

lim [ { (x^2 - sqrt(x))(x^2 + sqrt(x)) } / { (x^2 + sqrt(x)) sqrt(x - 1) } ]
x -> 1

lim [ (x^4 - x) / { (x^2 + sqrt(x)) sqrt(x - 1) } ]
x -> 1

Factoring out a sqrt(x) on the bottom, and an x on the top, we have

lim [ x(x^3 - 1) / { x^(1/2) (x^(3/2) + 1) sqrt(x - 1) } ]
x -> 1

Now we can cancel the x^(1/2) using the x on the top. Note that the x on the top is really x^1, so we can subtract exponents.

lim [ x^(1/2) (x^3 - 1) / { (x^(3/2) + 1) sqrt(x - 1) } ]
x -> 1

Let us also multiply top and bottom by sqrt(x - 1), to eliminate the radical on the bottom.

lim [ { x^(1/2) (x^3 - 1)sqrt(x - 1) } / { (x^(3/2) + 1) (x - 1) } ]
x -> 1

Now, we can factor x^3 - 1 as a difference of cubes.

lim [ { x^(1/2) (x - 1)(x^2 + x + 1)sqrt(x - 1) } / { (x^(3/2) + 1) (x - 1) } ]
x -> 1

And now, we can cancel out the (x - 1) term.

lim [ { x^(1/2) (x^2 + x + 1)sqrt(x - 1) } / { (x^(3/2) + 1) } ]
x -> 1

At this point, we can safely plug in the point x = 1, to give us

{ (1)^(1/2) (1^2 + 1 + 1)sqrt(1 - 1) } / { 1^(3/2) + 1 }

We can already see that sqrt(1 - 1) automatically gives us 0, so we get 0/2, which is equal to 0.

Therefore, our answer is 0.

Answer 3

Lt x tends to 1 x^2-sqrt x/sqrt x-1
=Lt x tends to 1 x(x-1)/sqrt x-1
=Lt x tends to 1 x(sqrt x-1*sqrt x+1)/sqrtx-1
=Lt x tends to 1 x(sqrt x+1)
PUT THE VALUE 1 IN X,
=1*sqrt2=sqrt2
Best of luck for the exams

Answer 4

If we put x=1 in the original function, it becomes 0/0, hence we can apply L'hospital rule

lim x->1 (x^2 - x^(0.5))/(x^(0.5) -1)
= lim x->1 (2*x - 0.5* *x^(-0.5))/(0.5 * x ^ (-0.5))
= (2 - 0.5)/0.5 (applying x =1 now as it is not 0/0 or y/0)
= 1.5/0.5
= 3

Answer 5

What does "sen" recommend? Do you recommend "sin"? i'm fairly constructive the reply is infinity, because the utmost degree contained in the numerator is two, even as the utmost degree contained in the denominator is a million (because the x² contained in the denominator is below an intensive, its degree is merely a million).

Answer 6

using l hospital rule , we get



0