After the widely read post Two Problems from IIT-JEE, I am going to discuss two problems which are not from IIT-JEE (as far as I am aware). They are taken from the masterpiece "A Course of Pure Mathematics"
by G. H. Hardy. The first one is a tough limit problem (at least I have
not been able to find a simpler solution till now) and the second one
is an instructive example which deals with the behavior of derivatives
for large values of the argument.
Problem 1: Find $ \displaystyle \lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$
I will try to use the simpler limit $ \displaystyle \lim_{x \to 0}\frac{\sin x - x}{x^{3}}$ in evaluating the above limit. For this simpler limit itself I have not been able to do away with the L'Hospital's rule. The best we can do without L'Hospital's rule (or Taylor's theorem) is that under the assumption that the limit exists we can prove that it has to be $ -1/6$.
So let's assume that $$ l = \lim_{x \to 0}\frac{\sin x - x}{x^{3}}$$ exists. Then we have \begin{align} l &= \lim_{x \to 0}\frac{\sin (3x) - 3x}{(3x)^{3}}\notag\\ &= \lim_{x \to 0}\frac{3\sin x - 4\sin^{3} x - 3x}{27 x^{3}}\notag\\ &= \frac{1}{9}\lim_{x \to 0}\frac{\sin x - x}{x^{3}} - \frac{4}{27}\lim_{x \to 0}\frac{\sin^{3}x}{x^{3}}\notag\\ &= \frac{l}{9} - \frac{4}{27}\notag\end{align} so that $ l = (9/8) \cdot (-4/27) = -1/6$.
But this does not establish that the limit $ l$ itself exists. For that we have to use the L'Hospital's rule to get $$\lim_{x \to 0}\frac{\sin x - x}{x^{3}} = \lim_{x \to 0}\frac{\cos x - 1}{3x^{2}} = \frac{1}{3}\cdot\left(-\frac{1}{2}\right) = -\frac{1}{6}$$ Clearly we also have $$\lim_{x \to 0}\frac{\sin x - x}{\sin^{3}x} = \lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\frac{x^{3}}{\sin^{3}x} = -\frac{1}{6}$$ Putting $ t = \sin x$ so that $ t \to 0$ as $ x \to 0$ we get $$\lim_{t \to 0}\frac{t - \sin^{-1}t}{t^{3}} = -\frac{1}{6}$$ Thus we have the following limits: $$\lim_{x \to 0}\frac{\sin x - x}{x^{3}} = -\frac{1}{6},\,\, \lim_{x \to 0}\frac{\sin^{-1}x - x}{x^{3}} = \frac{1}{6}$$
Multiplying these we get \begin{align} &\,\,\,\,\,\,\,\,\lim_{x \to 0}\frac{\{\sin x - x\}\{\sin^{-1}x - x\}}{x^{6}} = -\frac{1}{36}\notag\\ &\Rightarrow \lim_{x \to 0}\frac{\sin x\sin^{-1}x + x^{2} - x\{\sin x + \sin^{-1}x\}}{x^{6}} = -\frac{1}{36}\notag\\ &\Rightarrow \lim_{x \to 0}\frac{\sin x\sin^{-1}x - x^{2} + 2x^{2} - x\{\sin x + \sin^{-1}x\}}{x^{6}} = -\frac{1}{36}\notag\\ &\Rightarrow \lim_{x \to 0}\frac{\sin x\sin^{-1}x - x^{2}}{x^{6}} + \lim_{x \to 0}\frac{2x - \{\sin x + \sin^{-1}x\}}{x^{5}} = -\frac{1}{36}\notag \end{align} Putting $ x = \sin t$ in first limit on the left we get \begin{align} &\,\,\,\,\,\,\,\,\lim_{t \to 0}\frac{t\sin(\sin t) - \sin^{2}t}{\sin^{6}t} + \lim_{x \to 0}\frac{2x - \{\sin x + \sin^{-1}x\}}{x^{5}} = -\frac{1}{36}\notag\\ &\Rightarrow \lim_{t \to 0}\frac{t\sin(\sin t) - \sin^{2}t}{t^{6}}\cdot\frac{t^{6}}{\sin^{6}t} + \lim_{x \to 0}\frac{2x - \{\sin x + \sin^{-1}x\}}{x^{5}} = -\frac{1}{36}\notag\\ &\Rightarrow \lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}} + \lim_{x \to 0}\frac{2x - \{\sin x + \sin^{-1}x\}}{x^{5}} = -\frac{1}{36}\notag\\ &\Rightarrow A + B = -\frac{1}{36}\notag \end{align} where $ A$ is the limit to be calculated in the given problem. Clearly this will be done once we evaluate $ B$.
The limit $ B$ will be calculated using L'Hospital's Rule as follows: \begin{align} B &= \lim_{x \to 0}\frac{2x - \sin x - \sin^{-1}x}{x^{5}}\notag\\ &= \lim_{x \to 0}\dfrac{2 - \cos x - \dfrac{1}{\sqrt{1 - x^{2}}}}{5x^{4}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{5}\lim_{x \to 0}\dfrac{(2 - \cos x)\sqrt{1 - x^{2}} - 1}{x^{4}\sqrt{1 - x^{2}}}\notag\\ &= \frac{1}{5}\lim_{x \to 0}\dfrac{(2 - \cos x)\sqrt{1 - x^{2}} - 1}{x^{4}}\notag\\ &= \frac{1}{5}\lim_{x \to 0}\dfrac{(2 - \cos x)^{2}(1 - x^{2}) - 1}{x^{4}\{(2 - \cos x)\sqrt{1 - x^{2}} + 1\}}\notag\\ &= \frac{1}{10}\lim_{x \to 0}\dfrac{(1 + 2\sin^{2}(x/2))^{2}(1 - x^{2}) - 1}{x^{4}}\notag\\ &= \frac{1}{10}\lim_{t \to 0}\dfrac{(1 + 2\sin^{2}t)^{2}(1 - 4t^{2}) - 1}{16t^{4}}\notag\\ &= \frac{1}{160}\lim_{t \to 0}\dfrac{4\sin^{2}t + 4\sin^{4}t - 4t^{2} - 16t^{2}\sin^{2}t - 16t^{2}\sin^{4}t}{t^{4}}\notag\\ &= \frac{1}{160}\lim_{t \to 0}\left(4\dfrac{\sin^{2}t - t^{2}}{t^{4}} + 4\frac{\sin^{4}t}{t^{4}} - 16\frac{\sin^{2}t}{t^{2}} - 16t^{2}\frac{\sin^{4}t}{t^{4}}\right)\notag\\ &= \frac{1}{160}\lim_{t \to 0}\left(4\dfrac{\sin t - t}{t^{3}}\cdot\frac{\sin t + t}{t} + 4 - 16 - 0\right)\notag\\ &= \frac{1}{160}\left\{4\left(-\frac{1}{6}\right) 2 - 12\right\}\notag\\ &= -\frac{1}{12}\notag \end{align} So we have $ A = (-1/36) - B = (-1/36) + (1/12) = 1/18$. We therefore have $$\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}} = \frac{1}{18}$$ Note that in the above solution we have made use of L'Hospital's Rule twice. Second problem deals with the behavior of derivative of a function as its argument tends to infinity.
Problem 2: If $ f(x) + f'(x) \to a$ as $ x \to \infty$ then prove that $ f(x) \to a, f'(x) \to 0$ as $ x \to \infty$
Before we establish this it is simple to observe that if $ f(x)$ tends to a finite limit say $ l$ as $ x \to \infty$ and if the derivative $ f'(x)$ also tends to a limit then it must tend to zero. Clearly we have by the mean value theorem $$f(x) - f(x/2) = (x/2)f'(y)$$ where $ x/2 < y < x$ and we can see that the left hand side tends to zero and we don't have any option apart from the fact that the derivative should tend to zero.
Now back to original problem. Hardy argues in a very clever way. First we can reduce the problem to the case $ a = 0$ because if $ a \neq 0$ we can take $ f(x) = a + g(x)$ and then $ g(x) + g'(x) \to 0$. So let's assume that $ f(x) + f'(x) \to 0$ as $ x \to \infty$. If $ f'(x)$ remains positive for all large values of $ x$ then $ f(x)$ is increasing for all large values of $ x$ and hence it must tend to a limit $ l$ or to $ \infty$. If $ f(x) \to \infty$ then we must have $ f'(x) \to -\infty$ and this is contrary to the hypothesis that $ f'(x)$ is positive for large values of $ x$. If on the other hand $ f(x) \to l$ then we must have $ f'(x) \to -l$ and by what we have observed earlier in this case we must have $ l = 0$.
Similarly we can argue for the case when $ f'(x)$ is negative for all large values of $ x$. If $ f'(x)$ alternates its sign infinitely often for large values of $ x$ then it must vanish for infinitely many values of $ x$ (because derivatives possess the intermediate value property). Clearly these points will the maxima/minima of $ f$ and since $ f(x) + f'(x) \to 0$ and derivative vanishes at these points the function $ f$ must also be small at these points. Thus all the other values of the function $ f$ will be even smaller and therefore $ f(x) \to 0$ as $ x \to \infty$. And then clearly $ f'(x) \to 0$ as $ x \to \infty$.
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Problem 1: Find $ \displaystyle \lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$
I will try to use the simpler limit $ \displaystyle \lim_{x \to 0}\frac{\sin x - x}{x^{3}}$ in evaluating the above limit. For this simpler limit itself I have not been able to do away with the L'Hospital's rule. The best we can do without L'Hospital's rule (or Taylor's theorem) is that under the assumption that the limit exists we can prove that it has to be $ -1/6$.
So let's assume that $$ l = \lim_{x \to 0}\frac{\sin x - x}{x^{3}}$$ exists. Then we have \begin{align} l &= \lim_{x \to 0}\frac{\sin (3x) - 3x}{(3x)^{3}}\notag\\ &= \lim_{x \to 0}\frac{3\sin x - 4\sin^{3} x - 3x}{27 x^{3}}\notag\\ &= \frac{1}{9}\lim_{x \to 0}\frac{\sin x - x}{x^{3}} - \frac{4}{27}\lim_{x \to 0}\frac{\sin^{3}x}{x^{3}}\notag\\ &= \frac{l}{9} - \frac{4}{27}\notag\end{align} so that $ l = (9/8) \cdot (-4/27) = -1/6$.
But this does not establish that the limit $ l$ itself exists. For that we have to use the L'Hospital's rule to get $$\lim_{x \to 0}\frac{\sin x - x}{x^{3}} = \lim_{x \to 0}\frac{\cos x - 1}{3x^{2}} = \frac{1}{3}\cdot\left(-\frac{1}{2}\right) = -\frac{1}{6}$$ Clearly we also have $$\lim_{x \to 0}\frac{\sin x - x}{\sin^{3}x} = \lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\frac{x^{3}}{\sin^{3}x} = -\frac{1}{6}$$ Putting $ t = \sin x$ so that $ t \to 0$ as $ x \to 0$ we get $$\lim_{t \to 0}\frac{t - \sin^{-1}t}{t^{3}} = -\frac{1}{6}$$ Thus we have the following limits: $$\lim_{x \to 0}\frac{\sin x - x}{x^{3}} = -\frac{1}{6},\,\, \lim_{x \to 0}\frac{\sin^{-1}x - x}{x^{3}} = \frac{1}{6}$$
Multiplying these we get \begin{align} &\,\,\,\,\,\,\,\,\lim_{x \to 0}\frac{\{\sin x - x\}\{\sin^{-1}x - x\}}{x^{6}} = -\frac{1}{36}\notag\\ &\Rightarrow \lim_{x \to 0}\frac{\sin x\sin^{-1}x + x^{2} - x\{\sin x + \sin^{-1}x\}}{x^{6}} = -\frac{1}{36}\notag\\ &\Rightarrow \lim_{x \to 0}\frac{\sin x\sin^{-1}x - x^{2} + 2x^{2} - x\{\sin x + \sin^{-1}x\}}{x^{6}} = -\frac{1}{36}\notag\\ &\Rightarrow \lim_{x \to 0}\frac{\sin x\sin^{-1}x - x^{2}}{x^{6}} + \lim_{x \to 0}\frac{2x - \{\sin x + \sin^{-1}x\}}{x^{5}} = -\frac{1}{36}\notag \end{align} Putting $ x = \sin t$ in first limit on the left we get \begin{align} &\,\,\,\,\,\,\,\,\lim_{t \to 0}\frac{t\sin(\sin t) - \sin^{2}t}{\sin^{6}t} + \lim_{x \to 0}\frac{2x - \{\sin x + \sin^{-1}x\}}{x^{5}} = -\frac{1}{36}\notag\\ &\Rightarrow \lim_{t \to 0}\frac{t\sin(\sin t) - \sin^{2}t}{t^{6}}\cdot\frac{t^{6}}{\sin^{6}t} + \lim_{x \to 0}\frac{2x - \{\sin x + \sin^{-1}x\}}{x^{5}} = -\frac{1}{36}\notag\\ &\Rightarrow \lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}} + \lim_{x \to 0}\frac{2x - \{\sin x + \sin^{-1}x\}}{x^{5}} = -\frac{1}{36}\notag\\ &\Rightarrow A + B = -\frac{1}{36}\notag \end{align} where $ A$ is the limit to be calculated in the given problem. Clearly this will be done once we evaluate $ B$.
The limit $ B$ will be calculated using L'Hospital's Rule as follows: \begin{align} B &= \lim_{x \to 0}\frac{2x - \sin x - \sin^{-1}x}{x^{5}}\notag\\ &= \lim_{x \to 0}\dfrac{2 - \cos x - \dfrac{1}{\sqrt{1 - x^{2}}}}{5x^{4}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{5}\lim_{x \to 0}\dfrac{(2 - \cos x)\sqrt{1 - x^{2}} - 1}{x^{4}\sqrt{1 - x^{2}}}\notag\\ &= \frac{1}{5}\lim_{x \to 0}\dfrac{(2 - \cos x)\sqrt{1 - x^{2}} - 1}{x^{4}}\notag\\ &= \frac{1}{5}\lim_{x \to 0}\dfrac{(2 - \cos x)^{2}(1 - x^{2}) - 1}{x^{4}\{(2 - \cos x)\sqrt{1 - x^{2}} + 1\}}\notag\\ &= \frac{1}{10}\lim_{x \to 0}\dfrac{(1 + 2\sin^{2}(x/2))^{2}(1 - x^{2}) - 1}{x^{4}}\notag\\ &= \frac{1}{10}\lim_{t \to 0}\dfrac{(1 + 2\sin^{2}t)^{2}(1 - 4t^{2}) - 1}{16t^{4}}\notag\\ &= \frac{1}{160}\lim_{t \to 0}\dfrac{4\sin^{2}t + 4\sin^{4}t - 4t^{2} - 16t^{2}\sin^{2}t - 16t^{2}\sin^{4}t}{t^{4}}\notag\\ &= \frac{1}{160}\lim_{t \to 0}\left(4\dfrac{\sin^{2}t - t^{2}}{t^{4}} + 4\frac{\sin^{4}t}{t^{4}} - 16\frac{\sin^{2}t}{t^{2}} - 16t^{2}\frac{\sin^{4}t}{t^{4}}\right)\notag\\ &= \frac{1}{160}\lim_{t \to 0}\left(4\dfrac{\sin t - t}{t^{3}}\cdot\frac{\sin t + t}{t} + 4 - 16 - 0\right)\notag\\ &= \frac{1}{160}\left\{4\left(-\frac{1}{6}\right) 2 - 12\right\}\notag\\ &= -\frac{1}{12}\notag \end{align} So we have $ A = (-1/36) - B = (-1/36) + (1/12) = 1/18$. We therefore have $$\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}} = \frac{1}{18}$$ Note that in the above solution we have made use of L'Hospital's Rule twice. Second problem deals with the behavior of derivative of a function as its argument tends to infinity.
Problem 2: If $ f(x) + f'(x) \to a$ as $ x \to \infty$ then prove that $ f(x) \to a, f'(x) \to 0$ as $ x \to \infty$
Before we establish this it is simple to observe that if $ f(x)$ tends to a finite limit say $ l$ as $ x \to \infty$ and if the derivative $ f'(x)$ also tends to a limit then it must tend to zero. Clearly we have by the mean value theorem $$f(x) - f(x/2) = (x/2)f'(y)$$ where $ x/2 < y < x$ and we can see that the left hand side tends to zero and we don't have any option apart from the fact that the derivative should tend to zero.
Now back to original problem. Hardy argues in a very clever way. First we can reduce the problem to the case $ a = 0$ because if $ a \neq 0$ we can take $ f(x) = a + g(x)$ and then $ g(x) + g'(x) \to 0$. So let's assume that $ f(x) + f'(x) \to 0$ as $ x \to \infty$. If $ f'(x)$ remains positive for all large values of $ x$ then $ f(x)$ is increasing for all large values of $ x$ and hence it must tend to a limit $ l$ or to $ \infty$. If $ f(x) \to \infty$ then we must have $ f'(x) \to -\infty$ and this is contrary to the hypothesis that $ f'(x)$ is positive for large values of $ x$. If on the other hand $ f(x) \to l$ then we must have $ f'(x) \to -l$ and by what we have observed earlier in this case we must have $ l = 0$.
Similarly we can argue for the case when $ f'(x)$ is negative for all large values of $ x$. If $ f'(x)$ alternates its sign infinitely often for large values of $ x$ then it must vanish for infinitely many values of $ x$ (because derivatives possess the intermediate value property). Clearly these points will the maxima/minima of $ f$ and since $ f(x) + f'(x) \to 0$ and derivative vanishes at these points the function $ f$ must also be small at these points. Thus all the other values of the function $ f$ will be even smaller and therefore $ f(x) \to 0$ as $ x \to \infty$. And then clearly $ f'(x) \to 0$ as $ x \to \infty$.
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Readers should try the simpler (but similar) limit problem:
$\displaystyle \lim_{x \to 0}\frac{x\tan(\tan x) - \tan^{2}x}{x^{6}}$
Paramanand
March 20, 2013 at 9:52 AMFor Problem 2, consider $g(x)=\mathrm e^xf(x)$, note that $(a-\varepsilon)\mathrm e^x\leqslant g'(x)\leqslant(a+\varepsilon)\mathrm e^x$ for every $x$ large enough and integrate this. This yields $g(x)=(a\pm\varepsilon)\mathrm e^x+$ constant terms, for every $x$ large enough, hence $f(x)\to a$. QED.
Didier
March 20, 2013 at 10:00 AMThanks Didier for providing a different viewpoint based on integration. The argument is almost precise and rigorous except for the fact that during integration both the upper and lower limits must themselves be very large and then it will be difficult to reach the conclusion.Roughly the “+ constant terms” will actually not be constants but rather exponentials. The solution presented in the post does away with this problem.
Paramanand
March 20, 2013 at 10:01 AMNo. The lower limit is fixed and the upper limit goes to infinity hence the “+constant terms” will be, surprise… a constant term, which disappears in the limit when $x\to\infty$ since it is divided by $e^x$.
To sum up, you invented a problem where there is none.
did
September 23, 2013 at 4:08 PM@Did,
Thanks once again for correcting me. Somehow I had this thinking that the lower limit of integration is dependent on $x$ whereas in reality it is dependent on $\varepsilon$
Paramanand
September 23, 2013 at 5:05 PMI used taylor series and its manipulations to arrive at the answer in a fewer steps for the first question. The only question that would remain is the justification for the manipulation of the infinite series, which I'm sure can be given.
Sid
September 30, 2015 at 10:46 PM@Sid,
The Taylor series approach is very simple for the first problem. In fact with a little amount of manipulation you can see that the desired limit is equal to $$\lim_{x \to 0}\frac{\sin x\sin^{-1}x - x^{2}}{x^{6}}$$ and using the Taylor series for both $\sin x$ and $\sin^{-1}x$ the answer is arrived at very easily.
I consider Taylor series approach to be at a conceptually higher level than L'Hospital Rule (although the version of Taylor's Theorem used for evaluating limits can be proved using L'Hospital's Rule, see http://paramanands.blogspot.com/2013/11/teach-yourself-limits-in-8-hours-part-4.html) and hence prefer to avoid it for expository purposes.
Paramanand
October 1, 2015 at 9:14 AM