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I've heard the Laurent Series is the best way of dealing with most limits but I'm not sure if Laurent Series can help with this exact one.

$$\lim_{x\to\infty} \left( 2^{\large\sin\left(\tfrac{x^2+5}{x+5}\right)}-2^{\large\sin(x-5)} \right)=0$$

I believe the reason why it approaches zero is because as you graph it the amplitude becomes significantly lower until up to x=10,000 it is near zero.

I tried to use Laurent series, but I can't use when a=0. In my final step I substituted $\frac{1}{x}$ yet I couldn't isolate the variables to prevent the end result from being indeterminate for the Laurent series. It is also extremely difficult to use LHospital rule.

Is it possible to use the Laurent series at a=1 to solve the limit? Could you perhaps use the squeeze theorem?

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    $\begingroup$ That exponent is kind of hard to see. Do you want to put \large or \Large in front of that LaTeX? $\endgroup$ – Akiva Weinberger Apr 1 '15 at 15:17
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First, note that $$ 2^{\sin\left(\frac{x^2+5}{x+5}\right)}-2^{\sin(x-5)}=2^{\sin(x-5)}\left[2^{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}-1\right]. $$ Now, since $2^{\sin(x-5)}$ is bounded (it varies between $2$ and $1/2$), for the limit to be zero, $$ \lim_{x\rightarrow\infty}2^{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}-1 $$ must be $0$. In other words, $$ \lim_{x\rightarrow\infty}2^{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}=1 $$ Note that $$ \lim_{x\rightarrow\infty}2^{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}=2^{\lim_{x\rightarrow\infty}{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}}. $$ Therefore, you must show that $$ \lim_{x\rightarrow\infty}{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}=0. $$ Observe that $$ \lim_{x\rightarrow\infty}\frac{x^2+5}{x+5}-(x-5)=\lim_{x\rightarrow\infty}{\frac{x^2+5-(x^2-5)}{x-5}}=\lim_{x\rightarrow\infty}\frac{10}{x+5}. $$ Therefore, the difference between $\frac{x^2+5}{x+5}$ and $x-5$ approaches zero. By the mean value theorem, $$ \left(\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)\right)=\cos(c)\left|\frac{x^2+5}{x+5}-(x-5)\right| $$ where $c$ is between $\frac{x^2+5}{x+5}$ and $x-5$ (since $\cos(x)$ is the derivative of $\sin(x)$). Since $\cos(x)$ is a bounded function, $$ \lim_{x\rightarrow\infty}\left(\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)\right)=\lim_{x\rightarrow\infty}\cos(c)\left|\frac{x^2+5}{x+5}-(x-5)\right|\leq\lim_{x\rightarrow\infty}\left|\frac{x^2+5}{x+5}-(x-5)\right|=0. $$

This completes the computation.

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  • $\begingroup$ I could have, instead, just started with $g(y)=2^{\sin(y)}$ and looked at the MVT on this function, so you get $\lim_{x\rightarrow\infty}g'(c)|\frac{x^2+5}{x+5}-(x-5)|$ (where $c$ depends on $x$). The derivative is bounded, so we can skip to the last step. $\endgroup$ – Michael Burr Apr 1 '15 at 15:17

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