Question

If $\lim_{x\to 0} \frac{x+\sin x-x \cos x-\tan x}{x^{n}}$ exists and is non-zero finite value, then the value of n is

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (C)

5

Answer 2

To find the value of $n$, we need to analyze the Taylor series expansion of the numerator around $x=0$.

The Maclaurin series expansions for $\sin x$, $\cos x$, and $\tan x$ are:

1. $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + O(x^9)$

2. $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + O(x^8)$

3. $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + O(x^9)$

Substitute these expansions into the numerator $N(x) = x+\sin x-x \cos x-\tan x$:

$N(x) = x + \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \dots\right) - x\left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots\right) - \left(x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots\right)$

Expand the terms:

$N(x) = x + x - \frac{x^3}{6} + \frac{x^5}{120} - \dots - x + \frac{x^3}{2} - \frac{x^5}{24} + \dots - x - \frac{x^3}{3} - \frac{2x^5}{15} - \dots$

Group terms by powers of $x$:

Coefficient of $x$: $1 + 1 - 1 - 1 = 0$

Coefficient of $x^3$: $-\frac{1}{6} + \frac{1}{2} - \frac{1}{3} = \frac{-1 + 3 - 2}{6} = \frac{0}{6} = 0$

Coefficient of $x^5$: $\frac{1}{120} - \frac{1}{24} - \frac{2}{15} = \frac{1}{120} - \frac{5}{120} - \frac{16}{120} = \frac{1 - 5 - 16}{120} = \frac{-20}{120} = -\frac{1}{6}$

So, the numerator $N(x)$ simplifies to:

$N(x) = -\frac{x^5}{6} + O(x^7)$

Now, the limit becomes:

$\lim_{x\to 0} \frac{-\frac{x^5}{6} + O(x^7)}{x^{n}}$

For this limit to exist and be a non-zero finite value, the lowest power of $x$ in the numerator must match the power in the denominator. Therefore, $n$ must be 5.

In this case, the limit evaluates to:

$\lim_{x\to 0} \frac{-\frac{x^5}{6}}{x^{5}} = -\frac{1}{6}$

This is a non-zero finite value.

Thus, the value of $n$ is 5.