# Compute $\lim_{x \to 0} \frac{\sin(x)+\cos(x)-e^x}{\log(1+x^2)}$.

Question:

Compute $\lim_{x \to 0} \frac{\sin(x)+\cos(x)-e^x}{\log(1+x^2)}$.

Attempt: Using L'Hopital's Rule, I have come to $$\lim_{x \to 0} \frac{\cos(x)}{2x} - \lim_{x \to 0} \frac{\sin(x)}{2x} - \lim_{x \to 0} \frac{e^x}{2x}.$$ My thought was to use the power series representations of these functions. However, that' doesn't seem to get me anywhere. Am I on the wrong track using L'Hopital's?

• try using L'Hopital's rule again. Nov 27, 2014 at 20:39

Tricky computation: \begin{eqnarray*} \frac{\sin (x)+\cos (x)-e^{x}}{\log (1+x^{2})} &=&\frac{(\sin (x)-x)+(\cos (x)-1)-(e^{x}-1-x)}{\log (1+x^{2})} \\ &=&\frac{\frac{(\sin (x)-x)}{x^{2}}+\frac{(\cos (x)-1)}{x^{2}}-\frac{% (e^{x}-1-x)}{x^{2}}}{\frac{\log (1+x^{2})}{x^{2}}}. \end{eqnarray*} Now we use l'Hospital's rule to compute each of the four limits separatly

$\lim_{x\rightarrow 0}\frac{(\sin (x)-x)}{x^{2}}\overset{L^{\prime }HR}{=}% \lim_{x\rightarrow 0}\frac{(\cos (x)-1)}{2x}\overset{L^{\prime }HR}{=}% \lim_{x\rightarrow 0}\frac{-\sin (x)}{2}=\frac{-\sin (0)}{2}=0.$

$\lim_{x\rightarrow 0}\frac{(\cos (x)-1)}{x^{2}}\overset{L^{\prime }HR}{=}% \lim_{x\rightarrow 0}\frac{-\sin (x)}{2x}\overset{L^{\prime }HR}{=}% \lim_{x\rightarrow 0}\frac{-\cos (x)}{2}=\frac{-\cos (0)}{2}=-\frac{1}{2}$

$\lim_{x\rightarrow 0}\frac{(e^{x}-1-x)}{x^{2}}\overset{L^{\prime }HR}{=}% \lim_{x\rightarrow 0}\frac{e^{x}-1}{2x}\overset{L^{\prime }HR}{=}% \lim_{x\rightarrow 0}\frac{e^{x}}{2}=\frac{e^{0}}{2}=\frac{1}{2}$

$\lim_{x\rightarrow 0}\frac{\log (1+x^{2})}{x^{2}}\underset{x^{2}=y}{=}% \lim_{y\rightarrow 0^{+}}\frac{\log (1+y)}{y}\overset{L^{\prime }HR}{=}% \lim_{y\rightarrow 0^{+}}\frac{1/(1+y)}{1}=1.$

Therefore \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\sin (x)+\cos (x)-e^{x}}{\log (1+x^{2})} &=&\frac{% \lim_{x\rightarrow 0}\frac{(\sin (x)-x)}{x^{2}}+\lim_{x\rightarrow 0}\frac{% (\cos (x)-1)}{x^{2}}-\lim_{x\rightarrow 0}\frac{(e^{x}-1-x)}{x^{2}}}{% \lim_{x\rightarrow 0}\frac{\log (1+x^{2})}{x^{2}}} \\ &=&\frac{0+(-\frac{1}{2})-(\frac{1}{2})}{1}=-1. \end{eqnarray*}

• Tricky, indeed! This is a more explicit way of doing it than others have suggested, but the exposition is much appreciated. Nov 27, 2014 at 21:16
• you are welcome! Nov 27, 2014 at 21:20

Recall that, for $x$ near $0$, you have \begin{align} &e^x =1+x+\dfrac{x^2}{2}+\mathcal{O}(x^3)\\ &\cos x = 1-\dfrac{x^2}{2}+\mathcal{O}(x^3)\\ & \sin x = x+\mathcal{O}(x^3)\\ \end{align} giving $$\cos x+\sin x-e^x =- x^2+\mathcal{O}(x^3)$$ Since \begin{align} &\ln (1+ x^2) = x^2+\mathcal{o}(x^3) \end{align} then $$\lim_{x \to 0} \frac{\sin(x)+\cos(x)-e^x}{\log(1+x^2)}=\lim_{x \to 0} \frac{-x^2}{x^2}=-1.$$

• I'm not familiar with the Big-O and little-o notation in how you're using it, though I appreciate that your answer is different from the others. Can you explain the notation so I can understand better? Nov 27, 2014 at 21:12
• @flapjackery Please, have a look at this:math.columbia.edu/~nironi/taylor2.pdf Thank you. Nov 27, 2014 at 21:37
• Maybe in a near futur you will study a chapter in the infinite series course entitled ''Taylor expansion'' or Taylor series'' at that time it will be very clear what Olivia did. But if I have to give a short explanation, i tell you that in that course we learn how to find the nearest polynomiale to most functions, If it exists (of course). For example, the polynomial of degree 2 which is the much close to the function $e^x$ is $1+x+(x^2)/2$ and so on for the other functions, next we replace each function by the appropriate polynomial, the computation will simplifies and the limit is obtained. Nov 27, 2014 at 21:41

we get by the rule of L'Hospital $$\lim_{x \to 0}\frac{\cos(x)-\sin(x)-e^x}{\frac{2x}{1+x^2}}$$ $$\lim_{ x\to 0}\frac{(1+x^2)(\cos(x)-\sin(x)-e^x}{2x}$$ $$\lim_{x \to 0}\frac{2(\cos(x)-\sin(x)-e^x)+2x(-\sin(x)-\cos(x)-e^x)}{2}=0$$

• Thanks for your helpful answer. I think you may have made an error in the last step though, as I believe the limit equals -1, not 0. Nov 27, 2014 at 21:13
• One can also remove the $(1+x^2)$ factor from the numerator, because its limit is $1$. So the limit with the factor exists if and only if the limit without it exists and they're equal. However, the derivative of the numerator is completely wrong. Nov 27, 2014 at 21:30

Here's how I would remake the question: $$lim \left( {cos x \over 2x}-{e^x \over2x} \right)- lim {sinx \over 2x}$$ Use L'hopital's rule for the first and you should get negative infinity. The value of the second is equal to 1/2, so the limit is negative infinity.

• I think you may have misread the question. The limit should be $-1$. Though your suggestion in the comments was helpful, I did need to apply l'hopital's again. Nov 27, 2014 at 21:11

You made a mistake when you looked for derivative. When you apply L'Hopital's Rule you should get $$\lim_{x \to 0} \frac{\cos(x)-\sin(x)-e^x}{\frac{2x}{1+x^2}}$$ witch becomes $$\lim_{x \to 0} \frac{(x^2+1)(\cos(x)-\sin(x)-e^x)}{2x}$$ and when you apply L'Hopital's Rule again you should get $$\lim_{x \to 0} \frac{2x(\cos(x)-\sin(x)-e^x)+(1+x^2)(-\sin(x)-\cos(x)-e^x)}{2}=-1$$ Sorry for that mistake.

• You are using the product rule in the numerator to get to the last step, right? I think you might be missing a factor of $(1+x^2)$? Nov 27, 2014 at 21:14
• You're right! What I computed works for -cos x, not +. Sorry about that. Nov 27, 2014 at 21:40

Computation with l'Hospital's rule no trickly: \begin{eqnarray*} &&\lim_{x\rightarrow 0}\frac{\sin (x)+\cos (x)-e^{x}}{\log (1+x^{2})}\left. \overset{L^{\prime }HR}{=}\right. \lim_{x\rightarrow 0}\frac{\cos (x)-\sin (x)-e^{x}}{\frac{2x}{1+x^{2}}} \\ &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left. =\right. \lim_{x\rightarrow 0}\left( 1+x^{2}\right) \lim_{x\rightarrow 0}\left( \frac{\cos (x)-\sin (x)-e^{x}}{2x}\right) \end{eqnarray*} The first limit equals $+1.$ To compute the second limit we continue using l'Hospital's rule: $$\overset{L^{\prime }HR}{=}\lim_{x\rightarrow 0}\left( \frac{-\sin (x)-\cos (x)-e^{x}}{2}\right) =\frac{-0-1-1}{2}=-1.$$ Thereore $$\lim_{x\rightarrow 0}\frac{\sin (x)+\cos (x)-e^{x}}{\log (1+x^{2})}=1\times (-1)=-1.$$

Remark: If we have keeped the factor $1+x^{2}$ with the second one, the computation would be much more complicated; indeed, this factor has a contribution in complicating the computation. So putting it a side, it do note creat trouble!

If we have to keep it look the computation looks like this \begin{eqnarray*} &&\lim_{x\rightarrow 0}\frac{\sin (x)+\cos (x)-e^{x}}{\log (1+x^{2})}\left. \overset{L^{\prime }HR}{=}\right. \lim_{x\rightarrow 0}\frac{\cos (x)-\sin (x)-e^{x}}{\frac{2x}{1+x^{2}}} \\ &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left. =\right. \lim_{x\rightarrow 0}\frac{\left( 1+x^{2}\right) \left( \cos (x)-\sin (x)-e^{x}\right) }{2x} \\ &&\ \ \ \ \ \ \ \ \ \ \ \left. =\right. \lim_{x\rightarrow 0}\frac{\cos x-\sin x-e^{x}+x^{2}\cos x-x^{2}\sin x-x^{2}e^{x}}{2x} \\ &&\ \ \ \ \ \ \ \ \ \ \ \left. =\right. \lim_{x\rightarrow 0}\frac{-\sin x-\cos x-e^{x}-e^{x}x^{2}-x^{2}\sin x-x^{2}\cos x-2e^{x}x-2x\sin x+2x\cos x}{% 2} \\ &&\ \ \ \ \ \ \ \ \ \ \ \left. =\right. \frac{-0-1-1-0-0-0-0-0+0}{2} \\ &&\ \ \ \ \ \ \ \ \ \ \ \left. =\right. \frac{-2}{2} \\ &&\ \ \ \ \ \ \ \ \ \ \ \left. =\right. -1. \end{eqnarray*}