Ltx tends to 0
tan⁴x - sin⁴x / x⁶ =
a] 1
b]2
c]3
d]4

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{L}=\underset{\mathrm{x}\to 0}{\lim } \frac{{\tan }^4\mathrm{x}-{\sin }^4\mathrm{x}}{{\mathrm{x}}^6} \\ =\underset{\mathrm{x}\to 0}{\lim } \frac{{\displaystyle \frac{{\sin }^4\mathrm{x}}{{\cos }^4\mathrm{x}}}-{\sin }^4\mathrm{x}}{{\mathrm{x}}^6} \\ =\underset{\mathrm{x}\to 0}{\lim } \left(\frac{{\displaystyle \frac{1}{{\cos }^4\mathrm{x}}-1}}{{\mathrm{x}}^6}\right){\sin }^4\mathrm{x} \\ =\underset{\mathrm{x}\to 0}{\lim } \left(\frac{{\displaystyle 1-{\cos }^4\mathrm{x}}}{{\cos }^4\mathrm{x}.{\mathrm{x}}^6}\right){\sin }^4\mathrm{x} \\ \mathrm{Now} \underset{\mathrm{x}\to 0}{\lim } {\cos }^4\mathrm{x}=\cos 0=1 \\ \Rightarrow \mathrm{L}= \underset{\mathrm{x}\to 0}{\lim } \left\{\frac{{\displaystyle 1-{\left({\cos }^2\mathrm{x}\right)}^2}}{{\mathrm{x}}^6}\right\}{\sin }^4\mathrm{x} \\ = \underset{\mathrm{x}\to 0}{\lim } \left\{\frac{{\displaystyle \left(1-{\cos }^2\mathrm{x}\right)\left(1+{\cos }^2\mathrm{x}\right)}}{{\mathrm{x}}^6}\right\}{\sin }^4\mathrm{x} \\ =\underset{\mathrm{x}\to 0}{\lim } \left\{\frac{{\displaystyle {\sin }^2\mathrm{x}\left(1+{\cos }^2\mathrm{x}\right)}}{{\mathrm{x}}^6}\right\}{\sin }^4\mathrm{x} \\ =\underset{\mathrm{x}\to 0}{\lim } \left(1+{\cos }^2\mathrm{x}\right)\left(\frac{{\sin }^6\mathrm{x}}{{\mathrm{x}}^6}\right) \\ =\underset{\mathrm{x}\to 0}{\lim } \left(1+{\cos }^2\mathrm{x}\right){\left(\frac{\sin }{\mathrm{x}}\right)}^6 \\ \mathrm{We} \mathrm{know} \mathrm{that} \underset{\mathrm{x}\to 0}{\lim } \left(\frac{\sin }{\mathrm{x}}\right)=1 \\ \Rightarrow \mathrm{L}=\underset{\mathrm{x}\to 0}{\lim } \left(1+{\cos }^2\mathrm{x}\right) \\ =\left(1+{\cos }^{2}0\right) \\ =1+1 \\ =2 \\ \mathbf{Hence}\mathbf{ }\mathbf{option}\left(\mathbf{b}\right)\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{correct}\mathbf{.} \end{gathered}$$

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