Limit without L'Hopital or infinite series

Question

I have this limit which I had some troubles when trying to solve it $$ \lim_{x \to 0} \frac {e^x+e^{-x}-2}{\cos(4x)-\cos(x)} $$ If someone just gave me a hint.

Answer 1

Hint

$$\lim_{x \to 0}\frac {e^x+e^{-x}-2}{\cos(4x)-\cos(x)}=\lim_{x \to 0} \frac {(e^x-1)+(e^{-x}-1)}{-2\cdot\sin(5x/2)\sin(3x/2)}$$

Use Fundamental Limit:

$$\lim_{x \to 0}\frac {\sin x}{x}=1$$

$$\lim_{x \to 0}\frac {e^x-1}{x}=1$$

Answer 2

$$ \begin{align}\lim _{ x\to 0 } \frac { e^{ x }+e^{ -x }-2 }{ \cos (4x)-\cos (x) } &=\lim _{ x\to 0 } \frac { e^{ 2x }-2{ e }^{ x }+1 }{ \left( { e }^{ x } \right) \left( -2\sin { \frac { 3x }{ 2 } } \sin { \frac { 5x }{ 2 } } \right) } \\ &=\lim _{ x\to 0 } \frac { { \left( { e }^{ x }-1 \right) }^{ 2 } }{ \left( { e }^{ x } \right) \left( -2\sin { \frac { 3x }{ 2 } } \sin { \frac { 5x }{ 2 } } \right) } \\&=\lim _{ x\to 0 } \frac { { \left( { e }^{ x }-1 \right) }^{ 2 } }{ { x }^{ 2 } } \cdot \frac { \frac { 3x }{ 2 } \cdot \quad \frac { 5x }{ 2 } }{ \left( { e }^{ x } \right) \cdot \left( -2\sin { \frac { 3x }{ 2 } } \sin { \frac { 5x }{ 2 } } \right) } \cdot \frac { 2 }{ 3 } \cdot \frac { 2 }{ 5 } \\ &=-\frac { 2 }{ 15 } \end{align} $$

Answer 3

Hint $$\lim_{x \to 0} \frac{e^x+e^{-x}-2}{x^{2}}= \lim_{x \to 0} e^{-x}\frac{e^{2x}-2e^x+1}{x^{2}}= \lim_{x \to 0}e^{-x} \left( \frac{e^x-1}{x}\right)^2= \lim_{x \to 0}e^{-x} \left( \frac{e^x-e^0}{x-0}\right)^2$$ $$\lim_{x \to 0} \frac {\cos(4x)-\cos(x)}{x^2}=\lim_{x \to 0}-2 \frac {\sin(\frac{3x}{2})\sin(\frac{5x}{2})}{x^2}=-2\lim_{x \to 0} \frac {\sin(\frac{3x}{2})}{x}\frac {\sin(\frac{5x}{2})}{x} $$

Answer 4

Set $x=2t$, so the numerator is $$ e^{2t}+e^{-2t}-2=(e^t-e^{-t})^2 $$ The denominator is $$ \cos 4x-\cos x=\cos8t-\cos2t=-2\sin5t\sin3t $$ Thus the limit is $$ \lim_{t\to0}\frac{(e^t-e^{-t})^2}{-2\sin5t\sin3t}= \lim_{t\to0}-\frac{1}{2}\left(\frac{e^t-e^{-t}}{t}\right)^{\!2}\frac{t}{\sin5t}\frac{t}{\sin3t} $$ which you can surely compute.

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With l'Hôpital: $$ \lim_{x\to0}\frac{e^x+e^{-x}-2}{\cos 4x-\cos x}= \lim_{x\to0}\frac{e^x-e^{-x}}{\sin x-4\sin4x}= \lim_{x\to0}\frac{e^x+e^{-x}}{\cos x-16\cos4x} $$

With Taylor: $$ \lim_{x\to0}\frac{e^x+e^{-x}-2}{\cos 4x-\cos x}= \lim_{x\to0}\frac{1+x+\dfrac{x^2}{2}+1-x+\dfrac{x^2}{2}-2+o(x^2)} {1-\dfrac{(4x)^2}{2}-1+\dfrac{x^2}{2}+o(x^2)} $$

Answer 5

$e^x + e^{-x} - 2 = (e^{\frac {x}{2}} - e^{-\frac {x}{2}})^2 = (2\sinh \frac{x}{2})^2$

What about the denominator.

$\cos a x - \cos bx = \cos (\frac {a+b}{2} + \frac {a-b}{2}) x - \cos (\frac {a+b}{2} - \frac {a-b}{2})x = -2(\sin \frac{a+b}{2} x)(\sin \frac{a-b}{2} x)$

or in this case:

$-2(\sin \frac{5}{2} x)(\sin \frac{3}{2} x)$

$\lim_\limits {x\to 0} = \left(\frac {2 \sinh \frac {x}{2}}{x}\right)^2\left(\frac {x}{-2\sin \frac 52 x}\right)\left(\frac {x}{\sin \frac 32 x}\right) = \frac {1}{-2\cdot\frac {5}{2}\cdot\frac {3}{2}} = -\frac {2}{15}$

Answer 6

Just to give a different approach for dealing with the denominator, note that

$$(\cos4x-\cos x)(\cos4x+\cos x)=\cos^24x-\cos^2x=\sin^2x-\sin^24x=(\sin x-\sin4x)(\sin x+\sin4x)$$

and $\cos4x+\cos x\to2$ as $x\to0$, so that

$$\lim_{x\to0}{\cos4x-\cos x\over x^2}={1\over2}\left(\lim_{x\to0}{\sin x\over x}-\lim_{x\to0}{\sin4x\over x} \right)\left(\lim_{x\to0}{\sin x\over x}+\lim_{x\to0}{\sin4x\over x} \right)={1\over2}(1-4)(1+4)=-{15\over2}$$

Putting this together with

$${e^x-2+e^{-x}\over x^2}=\left(e^{x/2}-e^{-x/2}\over x\right)^2$$

and

$${e^{x/2}-e^{-x/2}\over x}={e^{x/2}-1\over x}-{e^{-x/2}-1\over x}\to{1\over2}-{-1\over2}={1\over2}+{1\over2}=1$$

gives the answer ${1^2\over-15/2}=-{2\over15}$.