Question

If $\underset{x\to 0}{lim}\left(\frac{3\sin x-x^3+\frac{x^3}{2}}{2x^n}\right)$ is a finite number, then greatest value of $n\in N$, is-

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is A $4$

$\underset{x\to 0}{lim}\left(\frac{3sinx-x^3+\frac{x^3}{2}}{2x^n}\right)$

Since this is of the form $\frac{0}{0}$

so, applying L- hospital rule,

$=\underset{x\to 0}{lim}\left(\frac{3cosx-\frac{3x^2}{2}}{2n{x}^{n-1}}\right)$

Again $\frac{0}{0}$ from, So by L-Hospital rule,

$=\underset{x\to 0}{lim}\frac{(-3sinx-\frac{6x}{2})}{2n(n-1)x^{x-2}}$

Again $\frac{0}{0}$ form, So by L-Hospital rule,

$=\underset{x\to 0}{lim}\left(\frac{-3cosx-3}{2n(n-1)(n-2)x^{n-3}}\right)$

Agin $\frac{0}{0}$ from

$=\underset{x\to 0}{lim}\left[\frac{3sinx}{2n(n-1)(n-2)(x-3)x^{x-4}}\right]$

$\because |\le sinx\le |$, so the above has the limit value if the denominate is equal to 1

$2n(n-1)(n-2)(n-3)x^{4-4}=1$

$\because 2n(n-1)(n-2)(n-3)\ne 1$;so,

$x^{n-4}=1=x^0\Rightarrow n-4=0$

$n=4$ So option (A) is correct