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Question: What is the problem in solving limit like this? $\lim_{x\to0} \frac{sinx-tanx}{x^3}$ $= \lim_{x\to...

What is the problem in solving limit like this?

lim⁡x→0sinx−tanxx3\lim_{x\to0} \frac{sinx-tanx}{x^3}

=lim⁡x→0sinxx.x2−tanxx.x2= \lim_{x\to0} \frac{sinx}{x.x^2} - \frac{tanx}{x.x^2}

=lim⁡x→01x2−1x2= \lim_{x\to0} \frac{1}{x^2} - \frac{1}{x^2}

=0= 0

The answer is not zero.

Answer

The problem is the incorrect application of limit properties for indeterminate forms.

Explanation

Solution

The problem in the given solution is the incorrect application of limit properties. The limit of the difference of two functions is equal to the difference of their limits only if the individual limits exist and are finite. In this case, lim⁡x→0sin⁡xx3=lim⁡x→0sin⁡xx⋅1x2=1⋅∞=∞\lim_{x\to0} \frac{\sin x}{x^3} = \lim_{x\to0} \frac{\sin x}{x} \cdot \frac{1}{x^2} = 1 \cdot \infty = \infty lim⁡x→0tan⁡xx3=lim⁡x→0tan⁡xx⋅1x2=1⋅∞=∞\lim_{x\to0} \frac{\tan x}{x^3} = \lim_{x\to0} \frac{\tan x}{x} \cdot \frac{1}{x^2} = 1 \cdot \infty = \infty The original limit is of the indeterminate form ∞−∞\infty - \infty. The step lim⁡x→0sinxx.x2−tanxx.x2=lim⁡x→01x2−1x2\lim_{x\to0} \frac{sinx}{x.x^2} - \frac{tanx}{x.x^2} = \lim_{x\to0} \frac{1}{x^2} - \frac{1}{x^2} incorrectly assumes that sin⁡xx\frac{\sin x}{x} and tan⁡xx\frac{\tan x}{x} can be replaced by their limit 1 while the remaining factor 1x2\frac{1}{x^2} tends to infinity. This is not allowed when the expression is in an indeterminate form. The approximations sin⁡x≈x\sin x \approx x and tan⁡x≈x\tan x \approx x are only the first terms in their Taylor series expansions, and for the given problem with x3x^3 in the denominator, higher order terms are significant.

To solve the limit correctly, we can use Taylor series expansions around x=0x=0: sin⁡x=x−x33!+O(x5)=x−x36+O(x5)\sin x = x - \frac{x^3}{3!} + O(x^5) = x - \frac{x^3}{6} + O(x^5) tan⁡x=x+x33+O(x5)\tan x = x + \frac{x^3}{3} + O(x^5) sin⁡x−tan⁡x=(x−x36)−(x+x33)+O(x5)=−x36−x33+O(x5)=−3x36+O(x5)=−x32+O(x5)\sin x - \tan x = (x - \frac{x^3}{6}) - (x + \frac{x^3}{3}) + O(x^5) = -\frac{x^3}{6} - \frac{x^3}{3} + O(x^5) = -\frac{3x^3}{6} + O(x^5) = -\frac{x^3}{2} + O(x^5) lim⁡x→0sin⁡x−tan⁡xx3=lim⁡x→0−x32+O(x5)x3=lim⁡x→0(−12+O(x2))=−12\lim_{x\to0} \frac{\sin x - \tan x}{x^3} = \lim_{x\to0} \frac{-\frac{x^3}{2} + O(x^5)}{x^3} = \lim_{x\to0} \left(-\frac{1}{2} + O(x^2)\right) = -\frac{1}{2}

Alternatively, we can manipulate the expression: lim⁡x→0sin⁡x−tan⁡xx3=lim⁡x→0sin⁡x(1−1cos⁡x)x3=lim⁡x→0sin⁡x(cos⁡x−1)x3cos⁡x\lim_{x\to0} \frac{\sin x - \tan x}{x^3} = \lim_{x\to0} \frac{\sin x (1 - \frac{1}{\cos x})}{x^3} = \lim_{x\to0} \frac{\sin x (\cos x - 1)}{x^3 \cos x} =lim⁡x→0sin⁡xx⋅cos⁡x−1x2⋅1cos⁡x= \lim_{x\to0} \frac{\sin x}{x} \cdot \frac{\cos x - 1}{x^2} \cdot \frac{1}{\cos x} We know that lim⁡x→0sin⁡xx=1\lim_{x\to0} \frac{\sin x}{x} = 1 and lim⁡x→01cos⁡x=11=1\lim_{x\to0} \frac{1}{\cos x} = \frac{1}{1} = 1. For lim⁡x→0cos⁡x−1x2\lim_{x\to0} \frac{\cos x - 1}{x^2}, we use the standard limit lim⁡x→01−cos⁡xx2=12\lim_{x\to0} \frac{1 - \cos x}{x^2} = \frac{1}{2}. So, lim⁡x→0cos⁡x−1x2=−lim⁡x→01−cos⁡xx2=−12\lim_{x\to0} \frac{\cos x - 1}{x^2} = -\lim_{x\to0} \frac{1 - \cos x}{x^2} = -\frac{1}{2}. Therefore, lim⁡x→0sin⁡x−tan⁡xx3=1⋅(−12)⋅1=−12\lim_{x\to0} \frac{\sin x - \tan x}{x^3} = 1 \cdot \left(-\frac{1}{2}\right) \cdot 1 = -\frac{1}{2}.

The problem in the given solution is the incorrect application of limit properties for indeterminate forms.