Question

Let functions $f$ and $g$ are defined as

$f(x) = \lim_{h\to0} \frac{\sin(x+3h)-3\sin(x+2h)+3\sin(x+h)-\sin x}{h(\tan(x+2h)-2\tan(x+h)+\tan x)}$ and

$g(x) = \frac{1-x(1+|1-x|)}{|1-x|} f(\frac{1}{1-x}) \sin \frac{1}{1-x} \sec^3 \frac{1}{1-x}$.

Then which of the following options is/are correct?

• A. $\lim_{x\to1^+} g(x)$ does not exists
• B. $\lim_{x\to1^-} g(x)$ does not exists
• C. $\lim_{x\to1^+} g(x) = 0$
• D. $\lim_{x\to1^-} g(x) = 0$

Answer 1

Answer: (A), (D)

(A), (D)

Answer 2

The function $f(x)$ is given by $f(x) = \lim_{h\to0} \frac{\sin(x+3h)-3\sin(x+2h)+3\sin(x+h)-\sin x}{h(\tan(x+2h)-2\tan(x+h)+\tan x)}$.

The numerator is the third forward difference of $\sin x$ with step size $h$, evaluated at $x$. Using the Taylor series expansion of $\sin(x+kh)$ around $h=0$, $\sin(x+kh) = \sin x + kh\cos x - \frac{(kh)^2}{2}\sin x - \frac{(kh)^3}{6}\cos x + O(h^4)$.

Numerator = $(\sin x + 3h\cos x - \frac{9h^2}{2}\sin x - \frac{27h^3}{6}\cos x) - 3(\sin x + 2h\cos x - \frac{4h^2}{2}\sin x - \frac{8h^3}{6}\cos x) + 3(\sin x + h\cos x - \frac{h^2}{2}\sin x - \frac{h^3}{6}\cos x) - \sin x + O(h^4)$ $= (1-3+3-1)\sin x + (3-6+3)h\cos x + (-\frac{9}{2}+\frac{12}{2}-\frac{3}{2})h^2\sin x + (-\frac{27}{6}+\frac{24}{6}-\frac{3}{6})h^3\cos x + O(h^4)$ $= 0 + 0 + 0 - \frac{6}{6}h^3\cos x + O(h^4) = -h^3\cos x + O(h^4)$.

The term in the denominator inside the parenthesis is the second forward difference of $\tan x$ with step size $h$, evaluated at $x$. Using the Taylor series expansion of $\tan(x+kh)$ around $h=0$, $\tan(x+kh) = \tan x + kh\sec^2 x + \frac{(kh)^2}{2}(2\sec^2 x \tan x) + O(h^3)$.

Denominator term = $(\tan x + 2h\sec^2 x + \frac{4h^2}{2}(2\sec^2 x \tan x)) - 2(\tan x + h\sec^2 x + \frac{h^2}{2}(2\sec^2 x \tan x)) + \tan x + O(h^3)$ $= (1-2+1)\tan x + (2-2)h\sec^2 x + (\frac{4}{2}-\frac{2}{2})h^2(2\sec^2 x \tan x) + O(h^3)$ $= 0 + 0 + h^2(2\sec^2 x \tan x) + O(h^3) = 2h^2\sec^2 x \tan x + O(h^3)$.

So, the denominator is $h(2h^2\sec^2 x \tan x + O(h^3)) = 2h^3\sec^2 x \tan x + O(h^4)$. $f(x) = \lim_{h\to0} \frac{-h^3\cos x + O(h^4)}{2h^3\sec^2 x \tan x + O(h^4)} = \frac{-\cos x}{2\sec^2 x \tan x} = \frac{-\cos x}{2 \frac{1}{\cos^2 x} \frac{\sin x}{\cos x}} = \frac{-\cos^4 x}{2\sin x}$.

Now consider $g(x) = \frac{1-x(1+|1-x|)}{|1-x|} f(\frac{1}{1-x}) \sin \frac{1}{1-x} \sec^3 \frac{1}{1-x}$.

Case 1: $x \to 1^+$. Then $x > 1$, so $1-x < 0$ and $|1-x| = -(1-x) = x-1$. $\frac{1-x(1+|1-x|)}{|1-x|} = \frac{1-x(1+x-1)}{x-1} = \frac{1-x^2}{x-1} = \frac{(1-x)(1+x)}{-(1-x)} = -(1+x)$ for $x \neq 1$.

Let $y = \frac{1}{1-x}$. As $x \to 1^+$, $1-x \to 0^-$, so $y \to -\infty$. $f(y) = \frac{-\cos^4 y}{2\sin y}$. $g(x) = -(1+x) \frac{-\cos^4 y}{2\sin y} \sin y \sec^3 y = (1+x) \frac{\cos^4 y}{2\sin y} \sin y \frac{1}{\cos^3 y} = (1+x) \frac{\cos y}{2}$.

$\lim_{x\to1^+} g(x) = \lim_{x\to1^+} (1+x) \frac{\cos(\frac{1}{1-x})}{2}$. Let $y = \frac{1}{1-x}$. As $x \to 1^+$, $y \to -\infty$. We need to evaluate $\lim_{y\to-\infty} (1+ (1-\frac{1}{y})) \frac{\cos y}{2} = \lim_{y\to-\infty} (2-\frac{1}{y}) \frac{\cos y}{2}$.

Since $\lim_{y\to-\infty} (2-\frac{1}{y}) = 2$ and $\cos y$ oscillates between -1 and 1 as $y \to -\infty$, $\lim_{y\to-\infty} \cos y$ does not exist. Therefore, $\lim_{y\to-\infty} (2-\frac{1}{y}) \frac{\cos y}{2}$ does not exist. Thus, $\lim_{x\to1^+} g(x)$ does not exist.

Case 2: $x \to 1^-$. Then $x < 1$, so $1-x > 0$ and $|1-x| = 1-x$. $\frac{1-x(1+|1-x|)}{|1-x|} = \frac{1-x(1+1-x)}{1-x} = \frac{1-x(2-x)}{1-x} = \frac{1-2x+x^2}{1-x} = \frac{(1-x)^2}{1-x} = 1-x$ for $x \neq 1$.

Let $y = \frac{1}{1-x}$. As $x \to 1^-$, $1-x \to 0^+$, so $y \to +\infty$. $f(y) = \frac{-\cos^4 y}{2\sin y}$. $g(x) = (1-x) \frac{-\cos^4 y}{2\sin y} \sin y \sec^3 y = (1-x) \frac{-\cos^4 y}{2} \frac{1}{\cos^3 y} = (1-x) \frac{-\cos y}{2}$.

$\lim_{x\to1^-} g(x) = \lim_{x\to1^-} (1-x) \frac{-\cos(\frac{1}{1-x})}{2}$. Let $y = \frac{1}{1-x}$. As $x \to 1^-$, $y \to +\infty$. We need to evaluate $\lim_{y\to+\infty} \frac{1}{y} \frac{-\cos y}{2} = \lim_{y\to+\infty} -\frac{\cos y}{2y}$.

Since $-1 \le \cos y \le 1$, we have $-\frac{1}{2y} \le -\frac{\cos y}{2y} \le \frac{1}{2y}$ for $y > 0$. As $y \to +\infty$, $\frac{1}{2y} \to 0$ and $-\frac{1}{2y} \to 0$. By the Squeeze Theorem, $\lim_{y\to+\infty} -\frac{\cos y}{2y} = 0$. Thus, $\lim_{x\to1^-} g(x) = 0$.

Based on the limits: (A) $\lim_{x\to1^+} g(x)$ does not exist. This is correct. (B) $\lim_{x\to1^-} g(x)$ does not exist. This is incorrect, the limit is 0. (C) $\lim_{x\to1^+} g(x) = 0$. This is incorrect, the limit does not exist. (D) $\lim_{x\to1^-} g(x) = 0$. This is correct.

The correct options are (A) and (D).