# Find Infinite limit (Rational Function) underset (x->5^+)(lim)(1/x^(4/3)-1/(x-5)^(4/3)) Question
Rational functions Find Infinite limit (Rational Function)
underset $$(x->5^+)(lim)(1/x^(4/3)-1/(x-5)^(4/3))$$ 2020-11-04
Given:
underset $$(x->5^+)(lim)(1/x^(4/3)-1/(x-5)^(4/3))$$
It is known that
underset $$(x->a)(lim)[f(x)+-g(x)]=underset(x->a)(lim)f(x)+-underset (x->a)(lim)g(x)$$
Then above can be written as
$$=underset(x->5+)(lim)(1/x^(4/3))-underset(x->5+)(lim)(1/(x-5)^4/3)$$
Plug Limits
$$(1/5^4/3)-(oo)(since,(x-5)^(4/3)>0)$$
$$-oo(since, c-oo=-oo)$$
Hence,
underset $$(x->5+)(lim)(1/x^(4/3)-1/(x-5)^4/3)=-oo$$

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$$f(x)=(2x^3+3x)/(9-7x^2)$$ The function $$\displaystyle{f{{\left({x}\right)}}}={x}\frac{{\left({64}-{x}^{{2}}\right)}^{{1}}}{{2}}$$ satisfies the hypotheses of Rolle's Theorem on the interval [-8,8]. Find all values of that satisfy the conclusion of the theorem.
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b.) $$\displaystyle+{4}\sqrt{{{2}}}{\quad\text{and}\quad}-{4}\sqrt{{{2}}}$$
c.) $$\displaystyle{4}\sqrt{{{2}}}$$
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My answer. The intervals do match and equal zero so Rolles theorem can work.
Second I found the derivative maybe thats where I can't solve this problem.
The derivative that I got was $$\displaystyle{64}-{x}^{{2}}+\frac{{x}}{\sqrt{{{64}-{x}^{{2}}}}}$$ maybe i did wrong on the simplifying. I at least tried hopefully some one can explain as much as possible with every single step because I can figure out the algebra part. Confirm that if x is rational and $$x \neq 0$$, then $$\frac{1}{x}$$ is rational.   