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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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