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## Homework Statement

Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}## without using L'Hospital's rule or expansion of the series.

Answer is given to be = ##\frac{(\log_e (2))^2}{2}##

## Homework Equations

Squeeze play theorem/ Sandwich theorem, some algebraic manipulations and standard results upto high school like

$$\lim_{x\to 0}\frac{a^x-1}{x} = \log_e (a)$$

Although we get the above formula by expansion but I think it would be valid to use it if we can. No I am not sure about it.

These are all the the methods that I can think of.

## The Attempt at a Solution

I have given multiple attempts to this question with a gap of several days in between, so I can't find all of them at this moment. I'll update the question when I find them. I have this one attempt with me at the time of posting this thread, so I'll write this down.

$$\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}$$

$$\rightarrow \lim_{x\to 0} \frac{e^{x\log_e 2}-1-x\log_e2}{x^2}$$

putting ##x\log_e2 = y##

$$\rightarrow (log_e 2)^2\lim_{y\to 0} \frac{e^{y}-1-y}{y^2}$$

$$\rightarrow (log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1+ye^{y/2}-y}{y^2}$$

$$\rightarrow (log_e 2)^2(\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2}+\lim_{y\to 0} \frac{ye^{y/2}-y}{y^2})$$

Solving it part by part

L

_{1}= $$(log_e 2)^2(\lim_{y\to 0} \frac{ye^{y/2}-y}{y^2})$$

$$\rightarrow (log_e 2)^2(\lim_{y\to 0} \frac{e^{y/2}-1}{2(y/2)}) = \frac{(log_e 2)^2}{2}$$

L

_{2}= $$(log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2}$$

for ## 1 \geqslant y \geqslant -1##

$$\rightarrow (log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2} \geqslant (log_e 2)^2\lim_{y\to 0} e^{y}-ye^{y/2}-1$$

$$\rightarrow (log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2} \geqslant 0 $$

That's it. I failed to find a function which can give ##\rightarrow (log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2} \leqslant 0 ## which would have completed the proof.

Any help would be greatly appreciated. Also if anyone can find a function which binds the limit as ##(log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2} \leqslant 0 ## , I would appreciate it if you can suggest some other ways than using sandwich theorem.