Mathematics - Limits Question with Solution | TestHub

Let $\underset{x\to 0}{\mathrm{Lt}}\frac{(\cos x-1)\left(\cos -e^n\right)}{n^x}=k$ (finite, non-zero)

$$\Rightarrow \underset{x\to 0}{\mathrm{Lt}}\frac{\left[\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\dots ..\right)-1\right]\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots .+\right)-\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}=1\right)}{x^n}=K$$

As the limit in finite, non zero we have degree of denominator = least power of x

$$\Rightarrow \mathrm{n}=3$$