Question

Solve the following limit:

(tanx - sinx)/(8x³) where x tends to 0.​

Answer 1

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

Given expression is

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{tanx - sinx}{ {8x}^{3} }$

If we substitute directly x = 0, we get

$\rm \:= \: \frac{tan0 - sin0}{ 0 }$

$\rm \:= \: \frac{0 - 0}{ 0 }$

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

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{tanx - sinx}{ {8x}^{3} }$

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ \dfrac{sinx}{cosx} - sinx}{ {8x}^{3} }$

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ \dfrac{sinx - sinx \: cosx}{cosx}}{ {8x}^{3} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \dfrac{sinx(1 - cosx)}{cosx \times {8x}^{3} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \dfrac{tanx(1 - cosx)}{{8x}^{3} }$

$\rm \: = \: \dfrac{1}{8} \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} \times \displaystyle\lim_{x \to 0}\rm \frac{1 - cosx}{ {x}^{2} }$

We know,

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

and

$\boxed{\sf{ \: \: 1 - cos2x = {2sin}^{2}x \: }}$

So, using this result, we get

$\rm \: = \: \dfrac{1}{8} \times 1 \times \displaystyle\lim_{x \to 0}\rm \frac{2 {sin}^{2} \dfrac{x}{2} }{ {x}^{2} }$

$\rm \: = \: \dfrac{1}{4} \displaystyle\lim_{x \to 0}\rm \frac{ {sin}^{2} \dfrac{x}{2} }{ {x}^{2} }$

$\rm \: = \: \dfrac{1}{4} \displaystyle\lim_{x \to 0}\rm \frac{ {sin} \dfrac{x}{2} \times {sin} \dfrac{x}{2} }{ \dfrac{x}{2} \times \dfrac{x}{2} \times 4}$

$\rm \: = \: \dfrac{1}{16} \displaystyle\lim_{x \to 0}\rm \frac{{sin} \dfrac{x}{2}}{ \dfrac{x}{2} } \times \displaystyle\lim_{x \to 0}\rm \frac{{sin} \dfrac{x}{2}}{ \dfrac{x}{2} }$

We know,

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

So, using this result, we get

$\rm \: = \: \dfrac{1}{16} \times 1 \times 1$

$\rm \: = \: \dfrac{1}{16}$

Hence,

$\rm\implies \: \:\boxed{\sf{ \: \: \rm \: \displaystyle\lim_{x \to 0}\rm \frac{tanx - sinx}{ {8x}^{3} } \: = \: \frac{1}{16} \: \: }}$

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Short Cut Trick :- Remember

$\boxed{\sf{ \: \:\displaystyle\lim_{x \to 0}\rm \frac{1 - cosx}{ {x}^{2} } = \frac{1}{2} \: \: }}$

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Additional Information :-

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

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} \: = \: 1 \: }}$

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} \: = \: loga \: }}$

Answer 2

• refer attachment for detailed answer