Question

If $\underset{x\to 1}{\mathop{\text{Lim}}}\,{{\left( 1+ax+{{x}^{2}} \right)}^{\dfrac{c}{x-1}}}={{e}^{3}}$ then, the value of bc is:
[a] 6
[b] -1
[c] 3
[d] -3

Answer 1

Hint: If $\underset{x\to a}{\mathop{\text{Lim}}}\,f(x)=0$ and $\underset{x\to a}{\mathop{\text{Lim}}}\,g(x)=0$ then $\underset{x\to a}{\mathop{\text{Lim}}}\,{{\left( 1+f(x) \right)}^{\dfrac{1}{g(x)}}}={{e}^{\underset{x\to a}{\mathop{\text{Lim}}}\,\dfrac{f\left( x \right)}{g\left( x \right)}}}$. Using this property and existence of the limit $\underset{x\to a}{\mathop{\text{Lim}}}\,\dfrac{f\left( x \right)}{g\left( x \right)}$, find two equations in a,b and c. Eliminate a to find the value of bc. Alternatively, you can use the L.Hospital rule to evaluate the limit $\underset{x\to a}{\mathop{\text{Lim}}}\,\dfrac{f\left( x \right)}{g\left( x \right)}$. Use the condition that the formed limit should be of $\dfrac{0}{0}or\dfrac{\infty }{\infty }$ or the denominator should be non-zero and finite. Hence form two equations in a,b and c. Eliminate a to find bc.

Complete step-by-step answer:
Since $\underset{x\to 1}{\mathop{\text{Lim}}}\,{{\left( 1+ax+{{x}^{2}} \right)}^{\dfrac{c}{x-1}}}$ exists, we have
$\begin{aligned} & \underset{x\to 1}{\mathop{\text{Lim}}}\,ax+b{{x}^{2}}=0 \\\ & \Rightarrow a+b=0\text{ (i)} \\\ \end{aligned}$

We know that If $\underset{x\to a}{\mathop{\text{Lim}}}\,f(x)=0$ and $\underset{x\to a}{\mathop{\text{Lim}}}\,g(x)=0$ then $\underset{x\to a}{\mathop{\text{Lim}}}\,{{\left( 1+f(x) \right)}^{\dfrac{1}{g(x)}}}={{e}^{\underset{x\to a}{\mathop{\text{Lim}}}\,\dfrac{f\left( x \right)}{g\left( x \right)}}}$.
Using the above result, we get
$\underset{x\to 1}{\mathop{\text{Lim}}}\,{{\left( 1+ax+{{x}^{2}} \right)}^{\dfrac{c}{x-1}}}={{e}^{\underset{x\to 1}{\mathop{\text{Lim}}}\,\dfrac{c\left( ax+b{{x}^{2}} \right)}{x-1}}}$

From equation (i), we have
a+b = 0
i.e. a = -b
Substituting the value of a, we get
$\underset{x\to 1}{\mathop{\text{Lim}}}\,{{\left( 1+ax+{{x}^{2}} \right)}^{\dfrac{c}{x-1}}}={{e}^{\underset{x\to 1}{\mathop{\text{Lim}}}\,\dfrac{c\left( -bx+b{{x}^{2}} \right)}{x-1}}}$

Since $\underset{x\to 1}{\mathop{\text{Lim}}}\,{{\left( 1+ax+{{x}^{2}} \right)}^{\dfrac{c}{x-1}}}={{e}^{3}}$, we have

& \underset{x\to 1}{\mathop{\lim }}\,\dfrac{c\left( -bx+b{{x}^{2}} \right)}{x-1}=3 \\\ & \Rightarrow \underset{x\to 1}{\mathop{\lim }}\,\dfrac{bcx\left( x-1 \right)}{x-1}=3 \\\ & \Rightarrow \underset{x\to 1}{\mathop{\lim }}\,bcx=3 \\\ & \Rightarrow bc=3 \\\ \end{aligned}$$ Hence the value of bc = 3. Hence option [c] is correct. Note: [1] LH rule: According to L. Hospital rule if $\underset{x\to a}{\mathop{\text{Lim} }}\,\dfrac{f\left( x \right)}{g\left( x \right)}$ is of the form $\dfrac{0}{0}or\dfrac{\infty }{\infty }$, then $\underset{x\to a}{\mathop{\text{Lim }}}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\text{Lim}}}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}$. Now, from above question, we have $\underset{x\to 1}{\mathop{\text{Lim }}}\,\dfrac{c\left( ax+b{{x}^{2}} \right)}{x-1}=3$ LHS should be of the form $\dfrac{0}{0}$. Hence we have a+b = 0 (i). Applying LH rule, we get $\begin{aligned} & \underset{x\to 1}{\mathop{\text{Lim }}}\,c\dfrac{a+2bx}{1}=3 \\\ & \Rightarrow c\left( a+2b \right)=3\text{ (ii)} \\\ \end{aligned}$ From equation (i), we have a+b = 0 i.e. a = -b Substituting in equation (ii), we get $\begin{aligned} & c\left( -b+2b \right)=3 \\\ & \Rightarrow bc=3 \\\ \end{aligned}$ Hence the value of bc is 3. Hence option [c] is correct.