Question

Find the value of $\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{(\sin x)(\cos x - \sin x - 1) + (\cos x)(\sin x + \cos x - 1)}}{x}} \right]$

Answer 1

Hint: In this question you need to find out the value of the given limit. Substitute in place of $x$ as 0, you will get $\dfrac{0}{0}$ form. Then use the L'Hospital rule that is differentiating numerator and denominator separately and substitute ‘x’ as 0 to get the answer.

Complete step-by-step answer:
Consider limit given,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{(\sin x)(\cos x - \sin x - 1) + (\cos x)(\sin x + \cos x - 1)}}{x}} \right]$---(1)
Now put $x = 0$
$ \Rightarrow \dfrac{{(\sin 0)(\cos 0 - \sin 0 - 1) + (\cos 0)(\sin 0 + \cos 0 - 1)}}{0}$
We know that $\sin 0 = 0$ and $\cos 0 = 1$
$ \Rightarrow \dfrac{{(0)(1 - 0 - 1) + 1(0 + 1 - 1)}}{0}$
$ \Rightarrow \dfrac{0}{0}$
Thus you will get $\dfrac{0}{0}$ form. If it is of $\dfrac{0}{0}$ form, then you need to apply L’hospital rule.
So, according to L’Hospital rule differentiate numerator and denominator separately with respect to $x$
Now applying L’Hospital rule for equation (1),
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\dfrac{d}{{dx}}(\sin x)(\cos x - \sin x - 1) + \dfrac{d}{{dx}}(\cos x)(\sin x + \cos x - 1)}}{{\dfrac{d}{{dx}}(x)}}} \right]$---(2)
Applying product rule for equation (2),
$Product\;rule = \left[ {\dfrac{d}{{dx}}(uv) = u\dfrac{d}{{dx}}(v) + v\dfrac{d}{{dx}}(u)} \right]$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\left[ {\sin x\dfrac{d}{{dx}}(\cos x - \sin x - 1) + (\cos x - \sin x - 1)\dfrac{d}{{dx}}(\sin x) + (\cos x)\dfrac{d}{{dx}}(\sin x + \cos x - 1) + (\sin x + \cos x - 1)\dfrac{d}{{dx}}(\cos x)} \right]}}{{\dfrac{d}{{dx}}(x)}}} \right]$
We know that,
$\dfrac{d}{{dx}}(1) = 0$, $\dfrac{d}{{dx}}(x) = 1$, $\dfrac{d}{{dx}}(\sin x) = \cos x$ and $\dfrac{d}{{dx}}(\cos x) = - \sin x$
After differentiating i.e., by using formula we get,
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\sin x( - \sin x - \cos x - 0) + (\cos x - \sin x - 1)(\cos x) + (\cos x)(\cos x - \sin x - 0) + ( - \sin x)(\sin x + \cos x - 1)} \right]$
On simplification we get,
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\sin x( - \sin x - \cos x) + (\cos x - \sin x - 1)\cos x + \cos x(\cos x - \sin x) - \sin x(\sin x + \cos x - 1)} \right]$
Multiply the terms and solve,
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ { - {{\sin }^2}x - \sin x\cos x + {{\cos }^2}x - \sin x\cos x - \cos x + {{\cos }^2}x - \sin x\cos x - {{\sin }^2}x - \sin x\cos x + \sin x} \right]$
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {2{{\cos }^2}x - 2{{\sin }^2}x - 4\sin x\cos x - \sin x - \cos x} \right]$
Now replace $x$ by 0 we get,
$ \Rightarrow 2{\cos ^2}(0) - 2{\sin ^2}(0) - 4\sin (0)\cos (0) - \sin (0) - \cos (0)$
$ \Rightarrow 2{(1)^2} - 2(0) - 4(0)(1) - 0 - 1$
$ \Rightarrow 2 - 1$
$ \Rightarrow 1$
Hence, the value is 1.

Note: To solve such questions check for $\dfrac{0}{0}$ form then only apply L’Hospital rule. And good understanding of the product rule of derivatives helps getting on the right track to reach the answer.