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Find

$\lim_{x\rightarrow 0}\frac{\exp{(ax)}-\exp{(bx)}+(b-a)x}{1-\cos{(x)}}$

using L'Hopital's Rule (and not Taylor).

As one said, using Taylor would probably make this problem easy, but the problem lies in fact that I cannot use taylor. Has anyone an idea?

Averroes2
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  • For a change, and after the semi fiasco of your previous, very recent, question, would you specify the tools you are allowed to use? And you definitely should explain what happened when you applied L'H to this limit. Because you did try something before posting, right? – Did Mar 05 '17 at 08:25
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    What are your ideas? – QFi Mar 05 '17 at 08:32

2 Answers2

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Using L'Hopital's rule twice works well: $$\lim_{x \to 0} \frac{e^{ax}-e^{bx}+(b-a)x}{1-\cos(x)}$$ $$=\lim_{x \to 0} \frac{ae^{ax}-be^{bx}+(b-a)}{\sin(x)}$$ $$=\lim_{x \to 0} \frac{a^2 e^{ax}-b^2 e^{bx}}{\cos(x)}$$ Can you continue?

projectilemotion
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$$e^{ax}-e^{bx}+(b-a)x=e^{ax}-ax-1-(e^{bx}-bx-1)$$

Now, using L'Hospital $$\lim_{x\to0}\dfrac{e^{cx}-cx-1}{1-\cos x}=c\lim_{x\to0}\dfrac{e^{cx}-1}{\sin x}=c^2\lim_{x\to0}\dfrac{e^{cx}}{\cos x}=c^2$$

Alternatively use Taylor's expansion, $e^{cx}=1+cx+\dfrac{(cx)^2}{2!}+\cdots$

lab bhattacharjee
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  • See this(http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion) the proof without L'Hôpital Rule or Series Expansion – lab bhattacharjee Mar 05 '17 at 10:51