Finding Limit using L'Hospital's Rule - Need Help!

[Yankel]

Hello,

I want to find this limit using L'hospital's rule. I got stuck after doing the derivaties.

\[\lim_{x\rightarrow \infty }\frac{xe^{\frac{x}{2}}}{x+e^{x}}\]

The answer should be 0, can you assist ? Thank you !

 

[evinda]

(Wave)

$$\lim_{x \to +\infty} \frac{xe^{\frac{x}{2}}}{x+e^x}=\lim_{x \to +\infty} \frac{(xe^{\frac{x}{2}})'}{(x+e^x)'}=\lim_{x \to +\infty} \frac{(x)'e^{\frac{x}{2}}+x(e^{\frac{x}{2}})'}{1+e^x}=\lim_{x \to +\infty} \frac{e^{\frac{x}{2}}+\frac{1}{2}x e^{\frac{x}{2}}}{1+e^x}=\lim_{x \to +\infty} \frac{\frac{1}{2}e^{\frac{x}{2}}+\frac{1}{2}e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}}}{e^x}=\lim_{x \to +\infty} \frac{1+\frac{1}{4}x}{e^{\frac{x}{2}}}=\lim_{x \to +\infty} \frac{\frac{1}{4}}{\frac{1}{2}e^{\frac{x}{2}}}=\lim_{x \to +\infty} \frac{2}{4} e^{-\frac{x}{2}}=0$$

 

[Nono713]

Here's another simple way to solve this limit. First recall the power series expansion of the exponential function is:
$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$
Which for $x > 0$ gives us:
$$\frac{e^x}{x^3} = \frac{1}{x^3} + \frac{1}{x^2} + \frac{1}{2x} + \frac{1}{6} + \cdots \geq \frac{1}{6}$$
Hence:
$$\frac{e^x}{x^3} \geq \frac{1}{6} ~ ~ ~ \iff ~ ~ ~ \frac{x^3}{e^x} \leq 6 ~ ~ ~ \iff ~ ~ ~ e^{-x} \leq \frac{6}{x^3} ~ ~ ~ \iff ~ ~ ~ e^{-\frac{x}{2}} \leq \frac{\sqrt{6}}{x \sqrt{x}}$$
Now for $x > 0$ we can squeeze the original expression as follows:
$$0 \leq \frac{xe^{\frac{x}{2}}}{x+e^{x}} \leq \frac{xe^{\frac{x}{2}}}{e^{x}} = x e^{- \frac{x}{2}}$$
And using the previous result:
$$0 \leq \frac{xe^{\frac{x}{2}}}{x+e^{x}} \leq x \frac{\sqrt{6}}{x \sqrt{x}} = \sqrt{\frac{6}{x}}$$
Clearly both the lower and upper bounds tend to zero as $x \to \infty$, so using the squeeze theorem we conclude:
$$\lim_{x \to \infty} \frac{xe^{\frac{x}{2}}}{x+e^{x}} = 0$$
The reason we consider $e^x / x^3$ is because $e^x / x^2$ isn't strong enough to show the limit is zero (only that it is finite). You can actually come up with arbitrarily tight upper bounds on your expression by considering higher and higher exponents, and the cubic (or, really, exponent $2 + \epsilon$) is the smallest one that actually show it tends to zero.​