How do I evaluate $$\lim_{x \to 0}\left(\frac{e^{2\sin x}-1}{x}\right)$$
I know it's the indeterminate form since the numerator and denominator both approach 0, but I can't use l'Hopital's rule so I'm not sure how to go about finding the limit.
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8 Answers
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Hint: Consider the function $f(x)=e^{2\sin x}$. What is the derivative of said function at $x=0$?
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$\begingroup$ The derivative would be $2\cos x e^{2sinx}$, which at 0 would be 2. But what about the rest of the function? $\endgroup$ Mar 16, 2015 at 1:47
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9$\begingroup$ But what is the definition of the derivative at $x=0$? $\endgroup$ Mar 16, 2015 at 1:50
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$\begingroup$ I'm not understanding this either (but seeing the upvotes, apparently other people do). Could you explain more? $\endgroup$ Mar 16, 2015 at 3:07
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6$\begingroup$ @user3932000: $f'(0) = \lim_{h \rightarrow 0} \frac{f(0 + h) - f(0)}{h}$. What happens when you set $f(x) = e^{2 \sin(x)}$ in that expression? What does the limit you get look like, after suitable simplification? $\endgroup$ Mar 16, 2015 at 5:28
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HINT: $\lim_{y\to0}\frac{e^y-1}{y}=1$. Hence $$ \lim_{x\to0}\frac{e^{2\sin x}-1}{x}= \lim_{x\to0}\frac{e^{2\sin x}-1}{2\sin x}\cdot\frac{2\sin x}{x} $$ and the rest should be easy.
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$\begingroup$ and how did you prove the hint? $\endgroup$ Mar 16, 2015 at 16:18
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$\begingroup$ @KonstantinosGaitanas First prove that $\lim_{y\to 0}\frac{\ln(y+1)}{y}=1$ ($\lim_{y\to 0}\ln\left( (1+y)^{\frac{1}{y}} \right)=\ln\lim_{y\to 0}\left( (1+y)^{\frac{1}{y}} \right)=\ln e = 1$). Now let $u=e^y -1$. Then $\lim_{y\to 0}\frac{e^y-1}{y}=\lim_{u\to 0}\frac{u}{\ln(u+1)}=1$. $\endgroup$ Mar 16, 2015 at 16:30
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$\begingroup$ @user314 and how do you prove the above limit you mention?Anyway I think that you have to use some form of L'Hospital or power series (which of course is a very difficult solution)to find the limit. $\endgroup$ Mar 16, 2015 at 16:36
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$\begingroup$ @KonstantinosGaitanas What do you mean? I proved $\lim_{y\to 0}\frac{\ln(y+1)}{y}=1$ in the comment if that's what you have in mind. $\endgroup$ Mar 16, 2015 at 16:37
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$\begingroup$ It's a widely known limit @konstantinos, the proof I know is based on the proof of a similar limit with the logarithmic function, that in turn it's based on the definition of $e$ as the limit of $\left(1+x\right)^\frac{1}{x}$ when $x\to 0$ $\endgroup$– seldonMar 16, 2015 at 16:58
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$$ \underbrace{f'(0) = \lim_{h\to0} \frac{f(0+h)-f(0)} h}_{\text{definition of ``derivative"}} = \lim_{h\to0}\frac{e^{2\sin h} - e^{2\sin 0}} h = \lim_{h\to0}\frac{e^{2\sin h} - 1} h. $$ So just find $f'(0)$ by the methods you would normally use to compute a derivative.
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A definition of the derivative is at $x=a$ is $$f'(a)=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$ Now plug in $a=0,f(x)=e^{2\sin x}$ and you'll have your answer.
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When given a task that includes the text "You are not allowed to use so-and-so Rule", you can still use the proof of that rule in your answer.
So, look up the proof of L'H in your text book or on Wikipedia, put in your function where apropriate. Ignore everything that isn't relevant to your case.
In the end you will have your answer as well as a better understanding of the rule.
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I guess the answer should be $2$. Multiply up and down by $2 \sin x$. Since we know $\lim_{x \to0} \frac{e^x -1}{x} =1$. Proceeding that way leaves us only with $\lim_{x \to0} \frac{2 \sin x}{x}$ (and we know $\lim_{x \to 0} \frac{\sin x}{x}=1$) So answer should be $2$.
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With $x$ in the neigborhood of $0$: $$e^{2\sin(x)}=1+2x+o(x^2)$$ so $$\frac{e^{2\sin(x)}-1}{x}=2+o(x)$$.
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Well, there is an easy solution if you know the series expansion of $e^x$.
Near $x=0$, $\sin(2x) \approx 2x$ therefor the given equation reduces to
$$\lim_{x\to0} \frac{e^{2x}-1}{x}$$
Now apply series expansion for $e^{2x}=1+x+...$ , neglecting higher powers as $x\to0$ There for it reduces to
$$\lim_{x\to0}\frac{1+x-1}{x}=1$$
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1$\begingroup$ I think you'll find that the series expansion for $e^{2x}$ is $1 + 2x + ...$, which gives a different answer... $\endgroup$ Mar 16, 2015 at 4:42
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