Problem with figuring out limit without using L'Hopital

Question

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?

Answer 1

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

Answer 2

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

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.)

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

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

Answer 3

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 $x\mapsto x^2$.

Answer 4

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,

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

Answer 5

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.