Question

Let $f(x)=3{{x}^{10}}-7{{x}^{8}}+5{{x}^{6}}-21{{x}^{3}}+3{{x}^{2}}-7$. The value of $\displaystyle \lim_{h \to 0}\dfrac{f(1-h)-f(1)}{{{h}^{3}}-3h}$ is equal to:
A. ${}^{22}/{}_{3}$
B. ${}^{50}/{}_{3}$
C. ${}^{25}/{}_{3}$
D. ${}^{53}/{}_{3}$

Answer 1

Observe the form of limit, i.e. find whether it is of indeterminate type or can be determined directly. Use the L'Hospital Rule to solve the expression. It states that if value of $\displaystyle \lim_{x \to c}\dfrac{f(x)}{g(x)}$ is of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, then we can differentiate the numerator and denominator individually and hence, can put limits to it. Use the formula:- $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$.

Complete step-by-step answer:
Given function in the problem is $f(x)=3{{x}^{10}}-7{{x}^{8}}+5{{x}^{6}}-21{{x}^{3}}+3{{x}^{2}}-7......(1)$
With the help of above function f(x), we need to determine the value of $\displaystyle \lim_{h \to 0}\dfrac{f(1-h)-f(1)}{{{h}^{3}}-3h}$.
So let the value of limit expression, i.e. $\displaystyle \lim_{h \to 0}\dfrac{f(1-h)-f(1)}{{{h}^{3}}-3h}$, be ‘M’.
It means, we can get an equation as
$M=\displaystyle \lim_{h \to 0}\dfrac{f(1-h)-f(1)}{{{h}^{3}}-3h}........(2)$
Now, we can put limit of $x\to 0$ to the equation (2), so we get value of M as,
$M=\displaystyle \lim_{h \to 0}\dfrac{f(1-h)-f(1)}{{{h}^{3}}-3h}=\dfrac{f(1)-f(1)}{0+0}=\dfrac{0}{0}$
$\therefore M=\dfrac{0}{0}.....(3)$.
It means the value of given limit in the problem is of indeterminate form.
So we can use L’Hospital Rule to determine the value of given limit i.e. value of M. L’Hospital Rule states that if the value of $\displaystyle \lim_{x \to c}\dfrac{f(x)}{g(x)}$ is of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, then we can differentiate f(x) and g(x) individually and hence try to put limit to the expression again, i.e. we get expression as $\displaystyle \lim_{x \to c}\dfrac{f'(x)}{g'(x)}$ if $\displaystyle \lim_{x \to c}\dfrac{f(x)}{g(x)}$ is of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$.
The same rule can be applied again, if we get indeterminate form of unit $\left( \dfrac{0}{0}~or\dfrac{\infty }{\infty } \right)$ again.
So we can differentiate numerator and denominator of the expression given in problem i.e. equation (2). So we can get the value of M as,
$M=\displaystyle \lim_{h \to 0}\dfrac{f(1-h)-f(1)}{{{h}^{3}}-3h}$
We know the identity $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$. So we get,

& M=\displaystyle \lim_{h \to 0}\dfrac{\dfrac{d}{dh}\left[ f(1-h)-f(1) \right]}{\dfrac{d}{dh}\left( {{h}^{3}}-3h \right)} \\\ & M=\displaystyle \lim_{h \to 0}\dfrac{\dfrac{d}{dh}f\left( 1-h \right)-\dfrac{d}{dh}f(1)}{\dfrac{d}{dh}{{h}^{3}}+\dfrac{d}{dh}\left( 3h \right)} \\\ \end{aligned}$$ We know $$\dfrac{d}{dx}\left( constant \right)=0$$. So we get the value of M as, $$\begin{aligned} & M=\displaystyle \lim_{h \to 0}\dfrac{-1f'\left( 1-h \right)-0}{3{{h}^{2}}+3\dfrac{d}{dh}\left( h \right)} \\\ & M=\displaystyle \lim_{h \to 0}\dfrac{-f'\left( 1-h \right)}{3{{h}^{2}}+3}.....(4) \\\ \end{aligned}$$ Now putting $$h\to 0$$ to the above expression, we get the value of M as, $$\begin{aligned} & M=\displaystyle \lim_{h \to 0}\dfrac{-f'\left( 1-0 \right)}{3{{\left( 0 \right)}^{2}}+3} \\\ & M=\displaystyle \lim_{h \to 0}\dfrac{-f'\left( 1 \right)}{3}......(5) \\\ \end{aligned}$$ As we are given the function f(x) in the problem or in equation (1), we can put the value of f’(1) in the equation (5) by finding the value of f’(x) and putting x = 1 to f’(x) expression. So, we get, $$f(x)=3{{x}^{10}}-7{{x}^{8}}+5{{x}^{6}}-21{{x}^{3}}+3{{x}^{2}}-7$$. Differentiate the above expression w.r.t. x with the help of the identity $$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$$ as, $$f'(x)=30{{x}^{9}}-56{{x}^{7}}+30{{x}^{5}}-63{{x}^{2}}+6x$$. Put x = 1 in the above expression to get the value of f’(1). So we get, $$\begin{aligned} & f'(x)=30{{(1)}^{9}}-56{{(1)}^{7}}+30{{(1)}^{5}}-63{{(1)}^{2}}+6(1) \\\ & f'(1)=30-56+30-63+6 \\\ & f'(1)=-53 \\\ \end{aligned}$$ Now, put value of f’(1) to the equation (5) to get the value of M. So we get, $$\begin{aligned} & M=\dfrac{-(-53)}{3} \\\ & M=\dfrac{53}{3} \\\ \end{aligned}$$ Hence value of $$\displaystyle \lim_{h \to 0}\dfrac{f(1-h)-f(1)}{{{h}^{3}}-3h}$$ is $$\dfrac{53}{3}$$. **So, the correct answer is “Option D”.** **Note:** Another approach for the question would be that we can find value of f(1) and f(1-h) from the given expression of f(x) directly and hence can put it to the given expression, i.e. $$\displaystyle \lim_{h \to 0}\dfrac{f(1-h)-f(1)}{{{h}^{3}}-3h}$$. So we get limit as $$f\left( 1-h \right)=3{{\left( 1-h \right)}^{10}}-7{{\left( 1-h \right)}^{8}}+5{{\left( 1-h \right)}^{6}}-21{{\left( 1-h \right)}^{3}}+3{{\left( 1-h \right)}^{2}}-7$$ $$f(1)=3{{\left( 1 \right)}^{10}}-7{{\left( 1 \right)}^{8}}+5{{\left( 1 \right)}^{6}}-21{{\left( 1 \right)}^{3}}+3{{\left( 1 \right)}^{2}}-7$$. Hence solve it with this approach as well. We cannot use L’Hospital property with other indeterminate form like $${{1}^{\infty }},{{0}^{\infty }},{{0}^{0}}$$ etc., we can use it only if the limit is of the form $$\dfrac{0}{0}$$ or $$\dfrac{\infty }{\infty }$$.