Math. Limits at Infinity?

Question

help me evaluate this. this is the only one I haven't answered

4) lim (sqrt1+sqrtx)/(sqrt1+x)
x-->+infinity
Note:
the numerator is a single term also the denominator

Answer 1

Are you saying that the expression inside the limit is
[ √(1+√x) ] / √(1+x) ?

If so, then you can alter this into:
√ [ (1+√x)/(1+x) ]
√ [ 1/(1+x) +√x/(1+x) ]
√ [ 1/(1+x) + 1/(1/√x + x/√x) ]
√ [ 1/(1+x) + 1/(1/√x + √x) ]

Now when you let x approach ∞, you get
√ [ 1/(1+∞) + 1/(1/√∞ + √∞) ]
√ [ 0 + 1/(0 + √∞) ]
√ [ 0 + 1/(√∞) ]
√ [ 0 + 0 ]
0

(Granted ∞ isn't a number you can really "plug in", but I used that notation above to show how and why the expression goes to 0)

Somebody said using l'Hopital's gives you a limit of 1, but the derivatives give you
[ (1/2)(1+√x)^(-1/2) (1/2)x^(-1/2) ] / (1/2)(1+x)^(-1/2)
[ (1+√x)^(-1/2) (1/2)x^(-1/2) ] / (1+x)^(-1/2)
2√(1+x) / √(1+√x) √x
2√(1+x) / √(√x + x)
You have to apply the rule again, which gives you

2(1/2)(1+x)^(-1/2) / (1/2)(√x + x)^(-1/2) *( (1/2)x^(-1/2) + 1)
4(1+x)^(-1/2) / (√x + x)^(-1/2) *( x^(-1/2) + 1)
Etc. I don't know. Seems like a mess to me, unless I made a mistake somewhere.

Answer 2

Answer: 0

sqrt1 =1 so the eqution is actually (1+sqrtx)/(1+x).
Divide the top and bottom by the highest value, i.e. x, getting
[1/x+sqrtx/x]/[1/x +1]. The denominator approaches 1 and the numerator reduces to 1/x +1/sqrtx, so it approaches 0 at infinity.

hope this helps

Answer 3

simplifying the equation, we got;

lim (1 + sqrt x)/(1 + x)
x--> + infinity

by just analyzing the numerator and denominator,

we can clearly see that the denominator (1+x) at any value +x will have a greater value than the numerator(1 + sqrt x);

thus;

the limit approaches simply ZERO.

Answer 4

lim (sqrt(1+sqrtx))/(sqrt(1+x)) <-- I think you mean this
x-->+infinity
The above gives infinity/infinity so use L'Hospital's Rule
which gives limit = 1