Finding the Limit of f(x)=(x+1)^(1/2)

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Homework Statement



lim f(x+h)-f(x)/h f(x)=(x+1)^(1/2) ans:1/2(x+1)^(1/2)
h→0

Homework Equations


The Attempt at a Solution


lim (x+h+1)^(1/2)-(x+1)^(1/2)/h lim (x+1)^(1/2)-(x+1)^(1/2)/h
h→0 h→0

Should the answer be D.N.E because the square roots cancel each other and leave 0/0
 
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\lim_{h \to 0} \frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}

Have you tried multiplying the numerator and denominator by the conjugate of the numerator?
 
Prototype44 said:

Homework Statement



lim f(x+h)-f(x)/h f(x)=(x+1)^(1/2) ans:1/2(x+1)^(1/2)
h→0

Homework Equations





The Attempt at a Solution


lim (x+h+1)^(1/2)-(x+1)^(1/2)/h lim (x+1)^(1/2)-(x+1)^(1/2)/h
h→0 h→0

Should the answer be D.N.E because the square roots cancel each other and leave 0/0
No. This type of limit always is of the [0/0] indeterminate form.
 
Figured it out
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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