# Problem with figuring out limit without using L'Hopital [closed]

I have two limits that I can't seem two solve without using L'Hopital's rule:

$$\lim _{x\to 2}\frac{2^{x}-x^{2}}{x-2}$$ and $$\lim _{x\to\pi/4}\frac{\sin{4x}\cos{2x}}{\left(x-\frac{\pi }{4}\right)^2}$$ Help anyone?

## closed as off-topic by heropup, Brevan Ellefsen, quid♦, Michael Albanese, colormegoneAug 9 '16 at 22:36

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• Thanks to the people here I think I got the first limit, but on the second one I tried using some basic trigonometric identities but I could never get the numerator to be anything else than 0. – intersomnium Aug 9 '16 at 16:32
• the numerator doesn't matter so much, if you still have 0/0 after differentiation, consider if a second differentiation will help, because you can tackle a further indeterminate form by using l'hopital again – Cato Aug 9 '16 at 16:37
• @AndrewDeighton I am not allowed to use L'Hopital.. – intersomnium Aug 9 '16 at 16:43
• silly me, I should have read it properly - oops – Cato Aug 9 '16 at 16:45

\begin{align} \lim_{x\to\pi/4}{\sin4x\cos2x\over(x-{\pi\over4})^2}&=\lim_{x\to\pi/4}{\sin(2\cdot2x)\cos2x\over({2x\over2}-{\pi\over4})^2}\\ &=\lim_{u\to\pi/2}{\sin2u\cos u\over({u\over2}-{\pi\over4})^2}\\ &=\lim_{u\to\pi/2}{8\sin u\cos^2u\over(u-{\pi\over2})^2}\\ &=\lim_{v\to0}{8\cos v\sin^2u\over v^2}\\ &=8(\lim_{v\to0}\cos v)\left(\lim_{v\to0}{\sin v\over v} \right)^2\\ &=8\cdot1\cdot1^2\\ &=8 \end{align}

• This would have never occurred to me. Thank you so much. :) – intersomnium Aug 9 '16 at 16:42
• the answers here don't use l'hopital though, for the first one you differentiate to (loge 2 2^x - 2x) / 1 = -1.227 when you plug 2 in, I'm not sure the answer below is 100% there – Cato Aug 9 '16 at 16:43
• @AndrewDeighton, did you mean to place your comment here or below the OP? – Barry Cipra Aug 9 '16 at 16:49
• @BarryCipra the moment I published the comment I remembered the trig identities. That's why I deleted the comment. But nevertheless thanks again hah. :) – intersomnium Aug 9 '16 at 19:16

$$\frac{2^x-x^2}{x-2}=\frac{2^x-4+4-x^2}{x-2}=4\frac{2^{x-2}-1}{x-2}-(x+2).$$

The second term tends to $-4$, while the first can be rewritten

$$\lim_{t\to0}\frac{2^t-1}t=\lim_{t\to0}\frac{e^{t\ln2}-1}t=\ln2\lim_{u\to0}\frac{e^{u}-1}u.$$

The last limit is known to be $1$. (For a justification of this fact, you need to provide your definition of the exponential.)

For the second limit, it is convenient to shift the variable,

$$\lim _{x\to\pi/4}\frac{\sin{4x}\cos{2x}}{\left(x-\frac{\pi }{4}\right)^2} = \lim _{t\to0}\frac{\sin{4t}\sin{2t}}{t^2} =\lim _{t\to0}4\frac{\sin{4t}}{4t}\lim _{t\to0}2\frac{\sin{2t}}{2t}.$$

• I understand everything up to the last point. How exactly did we get the last expression? :/ – intersomnium Aug 9 '16 at 16:14
• @intersomnium: $t\ln2=u$. – Yves Daoust Aug 9 '16 at 16:23

For the first limit, write $$\lim_{x\to 2} \frac{2^x -x^2}{x-2} = \lim_{x\to 2} \frac{2^x -4}{x-2} + \lim_{x\to 2} \frac{4-x^2}{x-2}.$$ Now use the difference quotient (or derivate) of $x\mapsto 2^x$ and $x\mapsto x^2$.

By using the formula $\color{red}{\sin(x-y)=\sin x\cos y-\cos x\sin y}$, and the fact that $\sin$ is an odd function we have $$\sin 4x =-(\sin 4x\cos \pi-\cos 4x\sin \pi)=-\sin(4x-\pi)=-\sin 4(x-\frac{\pi}4)\tag{1}$$ $$\cos 2x=\sin \frac{\pi}2\cos 2x-\cos\frac{\pi}2\sin 2x=\sin\left(\frac{\pi}2-2x\right)=-\sin 2\left(x-\frac{\pi}4\right)\tag{2}$$ So, putting $t=x-\frac{\pi}4$ we get

$$\lim_{x\to\frac{\pi}4}\frac{\sin 4x}{x-\frac{\pi}4}=-\lim_{x\to\frac{\pi}4}\frac{\sin 4(x-\frac{\pi}4)}{x-\frac{\pi}4}=-\lim_{t\to 0} \frac{\sin 4t}{t}=-4\lim_{t\to 0} \frac{\sin 4t}{4t}=-4\tag{1*}$$

$$\lim_{x\to\frac{\pi}4}\frac{\cos 2x}{x-\frac{\pi}4}=-\lim_{x\to\frac{\pi}4}\frac{\sin 2(x-\frac{\pi}4)}{x-\frac{\pi}4}=-\lim_{t\to 0} \frac{\sin 2t}{t}=-2\lim_{t\to 0} \frac{\sin 2t}{2t}=-2\tag{2*}$$

Then,

$$\lim _{x\to\pi/4}\frac{\sin{4x}\cos{2x}}{\left(x-\frac{\pi }{4}\right)^2}=\left(\lim_{x\to\frac{\pi}4}\frac{\sin{4x}}{x-\frac{\pi}4}\right)\left(\lim_{x\to\frac{\pi}4}\frac{\cos 2x}{x-\frac{\pi}4}\right)=(-4)(-2)=\boxed{\color{blue}{8}}$$

• @BarryCipra Yes, I had a mistake: taking $$\sin (4x-2\pi)=\sin 4\left(x-\frac{\pi}{4}\right)$$ Which is false, cause $$\sin 4\left(x-\frac{\pi}{4}\right)=\sin (4x-\pi)$$ Thanks for your observation. – Ángel Mario Gallegos Aug 9 '16 at 16:30

For the second one, let $f(x) = \sin 4x, g(x) = \cos 2x.$ Then the expression equals

$$\tag 1 \frac{(f(x) - f(\pi/4))(g(x) - g(\pi/4))}{(x-\pi)^2} = \frac{f(x) - f(\pi/4)}{x-\pi}\frac{g(x) - g(\pi/4)}{x-\pi}.$$

As $x\to \pi/4,$ $(1) \to f'(\pi/4)g'(\pi/4)$ from the definition of the derivative (no L'Hopital was used here). Now it's just a simple computation.