MCQ
$ \mathop {\lim }\limits_{\text{x} \to 0} \frac{{\sin 5\text{x}}}{{\tan 3\text{x}}}$
- A$ -\frac{5}{3}$
- B$ \frac{5}{3}$
- C$ -\frac{7}{3}$
- D$\text{None of these}$
Solution:
$ \mathop {\lim }\limits_{\text{x} \to 0} \frac{{\sin 5\text{x}}}{{\tan 3\text{x}}}$
$=\mathop {\lim }\limits_{\text{x} \to 0} \frac{{\sin 5\text{x}}}{{5\text{x}}}\times\frac{\text{3x}}{\tan\text{x}}\times\frac{5}{3}$
we know that $ =\displaystyle \lim_{\text{x}\rightarrow 0}\frac {\sin 5\text{x}}{5\text{x}}=1$
$=\mathop {\lim }\limits_{\text{x} \to 0}\times\frac{\text{3x}}{\tan\text{x}}=1$
$= \text{L}=1\times 1\times \dfrac {5}{3}$
$= \displaystyle \lim_{\text{x}\rightarrow 0}\frac {\sin 5\text{x}}{5\text{x}}=\frac{5}{3}$
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