Not So Rational Limit

Calculus Level 2

$\lim_{x \to 0}\frac{(x+4)^\frac{3}{2}+e^{x}-9}{x}$ Find the value of the closed form of the above limit.

The answer is 4.

1 solution

Amarildo Aliaj
Nov 9, 2016

First Method: L'Hopital $\lim _{x\to \:0}\left(\frac{\left(x+4\right)^{\frac{3}{2}}+e^x-9}{x}\right) =^H \lim _{x\to \:0}\left(\frac{3\sqrt{x+4}}{2}+e^x\right) = 4$

Second Method: Taylor $\left(x+4\right)^{\frac{3}{2}} \approx 8+3x+o(x^2)$ and $e^{x} \approx 1+x+o(x^2)$ $\lim_{x \to 0}\frac{(x+4)^\frac{3}{2}+e^{x}-9}{x} = \lim _{x\:\to \:0}\frac{8+3x+1+x-9}{x} = 4$

If you had to evaluate this without L'Hopital you can envision the definition of the derivative of f(x)=(x+4)^3/2-8 summed with the derivative of g(x)=e^x-1 at x=0. I know that in this scenario it is the exact same thing, but it is just a different way of looking at it.

Jonathan Russell - 2 years, 2 months ago

Not So Rational Limit

The answer is 4.

1 solution

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