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The value of $\underset{x \to 0}{\lim} \dfrac{\tan^{2}\:x-x\:\tan\:x}{\sin\:x}$ is

1. $\frac{\sqrt{3}}{2}$
2. $\frac{1}{2}$
3. $0$
4. None of these
in Calculus
edited | 42 views

Answer: $\mathbf C$

Solution:

Let $$\mathrm {I = \underset{x\to0}{\lim}\frac{\tan^2x - x\tan x}{\sin x}}$$

On substituting $\mathrm x = 0$, we can see that this integral is of the form, $\color {blue} {\mathbf{\frac{0}{0}}}$ i.e., $\color {blue} {\text {indeterminate form}}$

So, we can apply l'hospital's rule:

$$\therefore \mathrm {I = \underset{x\to0}{\lim}\frac{\frac{d}{dx}(\tan^2x-x\tan x)}{\frac{d}{dx}\sin x} }$$

$$\mathrm {\;\;\left(\because \frac{d}{dx}\tan x = \sec^2x,\; \frac{d}{dx}\cos x = \sin x\right)}$$

$$\mathrm {\Rightarrow \mathrm I = \underset{x \to 0}{\lim}\frac{\left (2\tan x \sec^2 x-(x.\sec^2x + 1.\tan x) \right)}{\cos x}}$$

$$\mathrm {\Rightarrow \mathrm I = \underset {x \to 0}{\lim}\left \{\left (2\frac{\sin x}{\cos x}.\frac{1}{\cos^2x} - \left(\frac{x}{\cos^2x}+\frac{\sin x}{\cos x}\right)\right)\right \}.\frac{1}{\cos x}}$$

Now, substitute $\mathrm x = 0$, we get:

$$\left (\text{Also}\;\because \cos 0 = 1, \;\sin 0 = 0\right)$$

$$\therefore \mathrm I = 0$$

Answer: $\therefore \mathbf C$ is the correct option.

by Boss (19k points)
edited by
+1

I think the pure mathematical notation for differential operator (with respect to $x$) is $\frac{d}{dx}()$ not $\frac{dy}{dx}$. So writing $\frac{dy}{dx}\sin x$ is incorrect. Rather we should write $\frac{d}{dx}(\sin x)$.

+1
Yes, correct.
+1
By the way, it was an inadvertent error from my side.

Thanks for pointing out such a minor but very important point.

PS: Corrected!