Question

Prove the following given limit.
$L=\displaystyle \lim_{x \to 1}\dfrac{\left( {{x}^{2}}+1 \right)\sin \left( x-1 \right)}{{{x}^{2}}-3x+2}=-2$

Answer 1

To solve this question we will first check that after applying the limit whether it is an indeterminate form or not, i.e. $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ is achieved or not. If so, then we will apply the L’Hospital rule by differentiating both the numerator and denominator separately with respect to x to get the result.

Complete step by step answer:
We are given,
$L=\displaystyle \lim_{x \to 1}\dfrac{\left( {{x}^{2}}+1 \right)\sin \left( x-1 \right)}{{{x}^{2}}-3x+2}$
We have to prove that its value is equal to – 2. We will consider the LHS of the equation.
$L=\displaystyle \lim_{x \to 1}\dfrac{\left( {{x}^{2}}+1 \right)\sin \left( x-1 \right)}{{{x}^{2}}-3x+2}$
Applying $\lim x \to 1$ we see that the above given limit is of the form $\dfrac{0}{0}$ which is one of the indeterminate forms. We solve this indeterminate form using the L’Hospital rule and hence differentiating both the numerator and denominator separately with respect to the given variable.
$L=\displaystyle \lim_{x \to 1}\dfrac{\left( {{x}^{2}}+1 \right)\sin \left( x-1 \right)}{{{x}^{2}}-3x+2}$
Differentiating both the numerator and denominator separately with respect to x and hence applying the L’Hospital rule, we get,
$L=\displaystyle \lim_{x \to 1}\dfrac{\dfrac{d}{dx}\left[ \left( {{x}^{2}}+1 \right)\sin \left( x-1 \right) \right]}{\dfrac{d}{dx}\left( {{x}^{2}}-3x+2 \right)}$
To differentiate the numerator, we will apply the product rule of differentiation stated as $\dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)=f\left( x \right)\dfrac{d}{dx}\left( g\left( x \right) \right)+g\left( x \right)\left( \dfrac{d}{dx}\left( f\left( x \right) \right) \right)$ where f(x) and g(x) are two functions of x. In the denominator, we will apply the formula of differentiation stated as $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}.$ Using both these rules to apply L’Hospital rule in L, we get,
$L=\displaystyle \lim_{x \to 1}\dfrac{\left( {{x}^{2}}+1 \right)\dfrac{d}{dx}\left( \sin \left( x-1 \right) \right)+\left[ \dfrac{d}{dx}\left( {{x}^{2}}+1 \right) \right]\sin \left( x-1 \right)}{\dfrac{d}{dx}\left( {{x}^{2}} \right)-\dfrac{d}{dx}\left( 3x \right)+\dfrac{d}{dx}\left( 2 \right)}$
$\Rightarrow L=\displaystyle \lim_{x \to 1}\dfrac{\left( {{x}^{2}}+1 \right)\cos \left( x-1 \right)+\left( 2x \right)\sin \left( x-1 \right)}{2x-3}$
Now, we observe that no $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form is achieved further. Hence we will directly apply $x \to 1$ on L, we get,
$\Rightarrow L=\dfrac{\left( 1+1 \right)\cos \left( 1-1 \right)+2\times 1\sin \left( 1-1 \right)}{2\times 1-3}$
$\Rightarrow L=\dfrac{2+2\times 0}{-1}$
$\Rightarrow L=-2$
This is obtained as sin 0 = 0 and cos 0 = 1.

Hence, we have proved that the value of $L=\displaystyle \lim_{x \to 1}\dfrac{\left( {{x}^{2}}+1 \right)\sin \left( x-1 \right)}{{{x}^{2}}-3x+2}$ is – 2.

Note: A key point to note here in this question is that we do not apply the L’Hospital rule in all the indeterminate form. Some of the indeterminate forms are given as $\dfrac{0}{0},\dfrac{\infty }{\infty },{{0}^{\infty }},0.\infty ,etc.$ We only apply L’Hospital rule in two indeterminate forms given by $\dfrac{0}{0}$ and $\dfrac{\infty }{\infty }.$ Also L’Hospital rule is stated as for functions f and g which are differentiable on an open interval I except possibly at a point c contained in I, if $\displaystyle \lim_{x \to c}f\left( x \right)=\displaystyle \lim_{x \to c}g\left( x \right)=0$ or $\pm \infty$ and ${{g}^{'}}\left( x \right)\ne 0\forall x\in I$ with $x\ne c$ and $\displaystyle \lim_{x \to c}\dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)}$ exists then $\displaystyle \lim_{x \to c}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to c}\dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)}.$