what is the limit?

Question

what is the limit as x-->1 (sqr(x)-1)/(x-1)

i just plugged in 1 in the denominator and got DNE but apparently thats not right so im wondering what the correct method is

Answer 1

lim ( √(x) - 1 ) / (x - 1)
x → 1

It isn't obvious, but you can factor the denominator as a difference of squares. Normally you cannot do this (because equations allow the possibility of x being both positive and negative numbers), but because this is a limit as x approaches 1, we can safely assume x is positive.

lim ( √(x) - 1 ) / [ (√(x) - 1) (√(x) + 1) ]
x → 1

Now, cancel the common factor.

lim 1 / [ (√(x) + 1) ]
x → 1

Now we can safely plug in x = 1.

1/[1 + 1]

1/2

Answer 2

The method is the L'opital rule which is derivative of the nominator over the derivative of denominator. In your case the lim = (1/(2*sqrt(x)))/1 and when you plug x = 1 your limit = 1/2.

Answer 3

The function y = (sqrt(x) - 1)/(x - 1) is not continuous at x = 1, so you cannot use direct substitution.

What you need to do is rationalize the numerator. Multiply both numerator and denominator of y by (sqrt(x) + 1).

lim x->1 ((sqrt(x) - 1)/(x - 1)
= lim x->1 [(sqrt(x) - 1)(sqrt(x) +1)]/[(x-1)(sqrt(x)+1)]
=lim x->1 (x-1)/[(x - 1)(sqrt(x) + 1)]
=lim x->1 (1/(sqrt(x) + 1))

Since y = 1/(sqrt(x) + 1) is continuous at x = 1, you can use direct substitution.

lim x->1 (1/(sqrt(x) + 1) = 1/(1 + 1) = 1/2

Answer 4

I suppose L'Hospital's Rule will work, but my approach would be to factor the denominator to (sqrt(x) + 1)(sqrt(x) - 1). Cancel out the (sqrt(x) - 1), and you are left with 1/(sqrt(x) +1). The limit of that is clearly 1/2.

Answer 5

Treat x-1 as the difference of two squares.

Remember
a²-b²=(a-b)(a+b)
so,
x-1=(Sqrt[x]-1)(Sqrt[x]+1)

The rest is pretty easy.

Answer 6

lim ( sqrt(x) -1)/x-1 (By l'Hospital's ruler)
x-->1

= lim 1/2sqrt(x)/1
x-->1

= lim 1/2(1)
x-->1

= lim 1/2
x-->1

= 1/2

Answer 7

lim(sqrt(x)-1)/x-1=lim(sqrt(x)-1)/{[(sqrt(x)-1][(sqrt(x)+1]}=
lim1/((sqrt(x)+1)=1/2