# Limit of a function tending to a finite number

If $$\lim_{x\to 0} \frac{ae^x - b\cos x +ce^{-x}}{x\sin x} = 2$$ then find the value of $a+b+c$.

My book has given the following solution to the above problem :-

We observe that as $x$ tends to zero , numerator tends to $a-b+c$ whereas the denominator tends to zero. Therefore for the limit to exist , we must have ,$a-b+c=0$

Now I am really confused at this point. Why would we want the numerator to attain the value of $0$ . Wouldn’t that give us an indeterminate answer? But actually it’s suposed to be two . Can you please explain ? Thank you for your help.

• If $a-b+c \neq 0$, the limit would be either $+\infty$ or $-\infty$. – Paolo Leonetti Mar 4 '18 at 10:33
• @Paolo but if $a-b+c=0$ then wouldn’t we get a $\frac{0}{0}$ form ? Which is anyways indeterminate – Aditi Mar 4 '18 at 10:37
• No. It is $0/0$, right, but it can have a limit depending on the values of $a,b,c$. – Paolo Leonetti Mar 4 '18 at 10:38

Hint. By using Taylor expansions at $0$, $$\frac{ae^x - b\cos x +ce^{-x}}{x\sin x} =\frac{a(1+x+\frac{x^2}{2}) - b(1-\frac{x^2}{2}) +c(1-x+\frac{x^2}{2})+o(x^2)}{x(x+o(x))}\\ =\frac{(a-b+c)+(a-c)x+\frac{(a+b+c)}{2}x^2+o(x^2)}{x^2+o(x^2)}.$$ In order to have the final limit $2$ we need that $$(a-b+c)=0,\quad(a-c)=0,\quad \frac{(a+b+c)}{2}=2.$$ Can you take it from here?

P.S. If you prefer to use L'Hopital begin with your initial observation $$\lim_{x\to 0}ae^x - b\cos x +ce^{-x}=a-b+c.$$ If $a-b+c=0$ then, by L'Hopital, $$\lim_{x\to 0}\frac{ae^x - b\cos x +ce^{-x}}{x}=\lim_{x\to 0}\frac{ae^x + b\sin x -ce^{-x}}{1}=a-c.$$ If $a-b+c=0$ AND $a-c=0$ then, by L'Hopital (twice), \begin{align} \lim_{x\to 0}\frac{ae^x - b\cos x +ce^{-x}}{x^2} &=\lim_{x\to 0}\frac{ae^x + b\sin x -ce^{-x}}{2x} \\&=\lim_{x\to 0}\frac{ae^x + b\cos x +ce^{-x}}{2}=\frac{a+b+c}{2}. \end{align}

• I’m sorry but I have just started learning this so I’m not very familiar with expansions – Aditi Mar 4 '18 at 10:38
• @Aditi If you don't like expansions, use de l'Hopital rule, you get the same conclusion – Paolo Leonetti Mar 4 '18 at 10:41
• Do you know L'Hopital's rule? – Robert Z Mar 4 '18 at 10:42
• Yes I’m familiar with that – Aditi Mar 4 '18 at 10:42
• So if I use L’Hospital’s rule and differentiate the numerator and denominator, will I get exactly the same conclusion ? – Aditi Mar 4 '18 at 10:43

If $a-b+c \neq 0$, the limit would be either $+\infty$ or $-\infty$.

Note, as written as Robert Z, that $$\frac{ae^x-b\cos x+cx^{-x}}{x\sin x}=\frac{(a-b+c)+x(a-c)+x^2(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}))+O(x^3)}{x^2+O(x^3)}$$ as $x\to 0$. Since the limit exists, you need $a-b+c=0$, $a=c$ and $\frac{a}{2}+\frac{b}{2}+\frac{c}{2}=2$. Therefore $$a+b+c=4$$ (and you also know that $a=1, b=2, c=1$).

• Thanks for helping ! I’ll try to familiarize myself with expansions so that I can get an idea of how to solve this properly :) – Aditi Mar 4 '18 at 10:56

If $a - b + c \ne 0$ then

$$\frac{ae^x-b\cos x+ce^{-x}}{x\sin x} \xrightarrow{x\to 0} \frac{a-b+c}{0} = \operatorname{sgn}(a-b+c) \cdot \infty \ne 2$$

Therefore $a - b + c = 0$. The limit is now an indeterminate form $\frac00$ so we can apply L'Hopital once. Differentiating gives

$$\frac{ae^x+b\sin x-ce^{-x}}{x\cos x+\sin x}$$

Again, if $a+b-c \ne 0$ then we would have:

$$\frac{ae^x+b\sin x-ce^{-x}}{x\cos x+\sin x} \xrightarrow{x\to 0} \frac{a+b-c}{0} = \operatorname{sgn}(a+b-c) \cdot \infty \ne 2$$

Therefore $a+b-c = 0$. The limit is now an indeterminate form $\frac00$ so we can again apply L'Hopital. Differentiating gives

$$\frac{ae^x+b\cos x+ce^{-x}}{-x\sin x+2\cos x}$$

Now the denominator is $\ne 0$ when $x\to 0$ so:

$$\frac{ae^x+b\cos x+ce^{-x}}{-x\sin x+2\cos x}\xrightarrow{x\to 0} \frac{a+b+c}{2}$$

This limit has to be $2$ so $$a+b+c = 4$$ $$a-b+c = 0$$ $$a+b-c = 0$$

The only solution is $a = 0, b = 2, c = 2$.

• by $sgn(a-b+c)$ do you mean the signum function ? – Aditi Mar 4 '18 at 10:57
• @Aditi Yes, it is just the sign of $a - b + c$. It only matters that it's $\ne 0$. – mechanodroid Mar 4 '18 at 10:58
• Okay thank you very much! – Aditi Mar 4 '18 at 11:06