Question

limit x tends to zero root(1+sin3x)-1/log (1+tan2x).​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\sf \frac{ \sqrt{1 + sin3x} - 1}{log(1 + tan2x)}$

If we substitute directly x = 0, we get

$\rm \: = \: \dfrac{ \sqrt{1 + sin0} - 1}{ log(1 + tan0) }$

$\rm \: = \: \dfrac{ \sqrt{1 + 0} - 1}{ log(1 + 0) }$

$\rm \: = \: \dfrac{ \sqrt{1} - 1}{ log(1) }$

$\rm \: = \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\sf \frac{ \sqrt{1 + sin3x} - 1}{log(1 + tan2x)}$

$\rm \: = \: \displaystyle\lim_{x \to 0}\sf ( \sqrt{1 + sin3x} - 1) \times \displaystyle\lim_{x \to 0}\sf \frac{1}{ log(1 + tan2x) }$

On rationalizing the numerator, we get

$\rm \: = \: \displaystyle\lim_{x \to 0}\sf ( \sqrt{1 + sin3x} - 1) \times \frac{ \sqrt{1 + sin3x} + 1}{ \sqrt{1 + sin3x} + 1} \times \displaystyle\lim_{x \to 0}\sf \frac{tan2x}{ log(1 + tan2x) } \times \frac{1}{tan2x}$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{ log(1 + x) }{x} = 1 \: }}}$

So, using this, we get

$\rm \: = \: \displaystyle\lim_{x \to 0}\sf \frac{1 + sin3x - 1}{ \sqrt{1 + sin3x} + 1} \times \displaystyle\lim_{x \to 0}\sf \frac{1}{tan2x}$

$\rm \: = \: \displaystyle\lim_{x \to 0}\sf \frac{ sin3x }{ \sqrt{1 + sin0} + 1} \times \displaystyle\lim_{x \to 0}\sf \frac{1}{tan2x}$

$\rm \: = \: \dfrac{ 1 }{ \sqrt{1 + 0} + 1} \times \displaystyle\lim_{x \to 0}\sf \frac{sin3x}{tan2x}$

$\rm \: = \: \dfrac{ 1 }{ \sqrt{1} + 1} \times \displaystyle\lim_{x \to 0}\sf \frac{sin3x}{3x} \times 3x \times \frac{2x}{tan2x} \times \frac{1}{2x}$

$\rm \: = \: \dfrac{ 1 }{ 1 + 1} \times \displaystyle\lim_{x \to 0}\sf \frac{sin3x}{3x} \times 3 \times \frac{2x}{tan2x} \times \frac{1}{2}$

We know

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{sinx}{x} = 1 \: }}}$

and

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{tanx}{x} = 1 \: }}}$

So, using these results, we get

$\rm \: = \: \dfrac{3}{4} \times 1 \times 1$

$\rm \: = \: \dfrac{3}{4}$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{ \sqrt{1 + sin3x} - 1}{log(1 + tan2x)} = \frac{3}{4} \: }}$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$