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valentinak56 [21]
3 years ago
8

Because of their connection with secant​ lines, tangents, and instantaneous​ rates, limits of the form ModifyingBelow lim With h

right arrow 0 StartFraction f (x plus h )minus f (x )Over h EndFraction occur frequently in calculus. Evaluate this limit for the given value of x and function f. f (x )equals StartRoot x EndRoot​, xequals11
Mathematics
1 answer:
Gre4nikov [31]3 years ago
8 0

Answer:

\dfrac{1}{2\sqrt{x}}

Step-by-step explanation:

f(x) = \sqrt{x} = x^{\frac{1}{2}}

f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}

We use binomial expansion for (x+h)^{\frac{1}{2}}

This can be rewritten as

[x(1+\dfrac{h}{x})]^{\frac{1}{2}}

x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}

From the expansion

(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots

Setting x=\dfrac{h}{x} and n=\frac{1}{2},

(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots

=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots

Multiplying by x^{\frac{1}{2}},

x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots

x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots

\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots

The limit of this as h\to 0 is

\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}} (since all the other terms involve h and vanish to 0.)

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