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Jun 2022
By birnial -   In Uncategorized -   Comments Off on [FSX] Thai Creation – Nepal, Kathmandu VNKT (FSX Portover) Download

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[FSX] Thai Creation – Nepal, Kathmandu VNKT (FSX Portover) Download

 

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Category:Airports in Nepal
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Category:1958 establishments in NepalQ:

Find the limit $\lim_{x \to 0}\frac{e^x-\frac{1}{x}}{x}$

Question – I need to find the limit
$$\lim_{x \to 0}\frac{e^x-\frac{1}{x}}{x}$$
Here’s how I attempted it:
$$e^x = \lim_{x \to 0}\frac{e^x-1}{x-0} = \lim_{x \to 0}\frac{e^x}{x} = 1$$
$$\frac{1}{x} = \lim_{x \to 0}\frac{1}{x-0} = \lim_{x \to 0}\frac{1}{x} = 0$$
$$\lim_{x \to 0}\frac{e^x-\frac{1}{x}}{x}= \lim_{x \to 0}\frac{e^x-0}{x-0} = \lim_{x \to 0}\frac{e^x}{x}$$
I know the answer is $\frac{1}{2}$, but how to I reach that conclusion?

A:

Hint: Let $f(x)=e^x-\frac{1}{x}$. Then $f'(x)=e^x\cdot \ln x-\frac{1}{x^2}$ and hence $f'(x)\to \infty$ as $x\to 0$, so $f$ is unbounded near $0$. Hence
$$
\lim_{x\to 0}\frac{e^x-\frac{1}{x}}{x}=\lim_{x\to 0}\frac{f(x)}{f'(x)}=\ https://atlantickneerestoration.com/simple-map-crack-full-version-pc-windows-updated-2022/

 

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