- #1
Umrao
- 31
- 6
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
L1 = $$(log_e 2)^2(\lim_{y\to 0} \frac{ye^{y/2}-y}{y^2})$$
$$\rightarrow (log_e 2)^2(\lim_{y\to 0} \frac{e^{y/2}-1}{2(y/2)}) = \frac{(log_e 2)^2}{2}$$
L2 = $$(log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2}$$
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.