Solve without the use of L'Hopital
$$\lim\limits_{x\to1}{\frac{\sqrt{2^x+7}-\sqrt{2^{x+1}+5}}{x^3-1}}$$
I multiplied by the conjugate but nothing has changed.
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2Use L'Hospital rule – Empty Dec 03 '15 at 16:40
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use l hospital rule – Dec 03 '15 at 16:40
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solved without the use of L'Hopital – Olya12 Dec 03 '15 at 16:42
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4"Nothing happened"? Did you wait for a while for something to happen? – Dec 03 '15 at 16:44
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1After you do the conjugate, we basically want the limit of $\frac{2+2^x-2^{x+1}}{x-1}$. (There is an easy constant involved.) To make it look nicer, change the sign of the top. Let $f(x)=2^{x+1}-2^x$. Then by the definition of the derivative we want $f'(1)$. – André Nicolas Dec 03 '15 at 17:12
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the denominator is always zero – Olya12 Dec 03 '15 at 17:15
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When you say "solved without the use of L'Hopital", do you mean that you are required to solve it without L'Hopital, or that you have not tried L'Hopital yet? – Brian Tung Dec 03 '15 at 18:23
1 Answers
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We can evaluate the limit of interest by using only well-known bounds for the exponential function along with the Squeeze Theorem. To that end we proceed.
Note that we can write
$$\begin{align} \frac{\sqrt{2^x+7}-\sqrt{2^{x+1}+5}}{x^3-1}&=\frac{2-2^x}{\left(\sqrt{2^x+7}+\sqrt{2^{x+1}+5}\right)(x^3-1)}\\\\ &=\left(\frac{-2}{\left(\sqrt{2^x+7}+\sqrt{2^{x+1}+5}\right)(x^2+x+1)}\right)\left(\frac{2^{x-1}-1}{x-1}\right)\\\\ &=\left(\frac{-2}{\left(\sqrt{2^x+7}+\sqrt{2^{x+1}+5}\right)(x^2+x+1)}\right)\left(\frac{e^{(x-1)\log 2}-1}{x-1}\right)\\\\ \end{align}$$
Now, in THIS ANSWER, I showed that for $|x|<1$, the exponential function satisfies the inequalities
$$1+x \le e^x\le \frac{1}{1-x}$$
Therefore, we have for $x<1+\frac{1}{\log 2}$
$$\log 2 \le \frac{e^{(x-1)\log 2}-1}{x-1}\le \frac{\log 2}{1-(x-1)\log 2}$$
whereupon invoking the Squeeze Theorem gives
$$\lim_{x\to 1} \frac{e^{(x-1)\log 2}-1}{x-1}=\log 2$$
And finally, we have
$$\lim_{x\to 1}\frac{\sqrt{2^x+7}-\sqrt{2^{x+1}+5}}{x^3-1}=-\frac19 \log 2$$
Mark Viola
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