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When \lim _{x \rightarrow-1} \frac{x^2-a x+3 b}{x+1}=8 , then which of the following is true?

Option: 1

a \leq b


Option: 2

a< b


Option: 3

a> b


Option: 4

Cannot be determined


Answers (1)

best_answer

Apply the L' Hospital's Rule, when  \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{0}{0}, \frac{\infty}{\infty}, \text { etc. }  [an intermediate form] in the following way

  • Differentiate \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)} , provided that the limit   \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}  exists.

  • Otherwise, go on differentiating till the limit with the determinate form is achieved.

The provided limit is

\lim _{x \rightarrow-1} \frac{x^2-a x+3 b}{x+1}=8

Here use the L’ Hospital rule and also refer to the following.

  • When x\rightarrow -1 , it implies (x+1)\rightarrow 0 .

  • The limit will take the form of  \frac{0}{0} , when x\rightarrow -1 .

Now, derive the following: 

\begin{aligned} & \left.\left(x^2-a x+3 b\right)\right|_{s=-1}=0 \\ & (-1)^2-a(-1)+3 b=0 \\ & 1+a+3 b=0 \\ & -a=1+3 b \quad \ldots(i) \end{aligned}

Use the equation (i) to calculate the following:

\begin{aligned} & x^2-a x+3 b \\ & =x^2+(1+3 b) x+3 b \\ & =x^2+x+3 b x+3 b \\ & =x(x+1)+3 b(x+1) \\ & \therefore x^2-a x+3 b=(x+1)(x+3 b) \end{aligned}  ...(ii)

Use the equation (ii) to rewrite the limit.

\begin{aligned} & \lim _{x \rightarrow-1} \frac{x^2-a x+3 b}{x+1}=8 \\ & \lim _{x \rightarrow-1} \frac{(x+1)(x+3 b)}{(x+1)}=8 \\ & \lim _{x \rightarrow-1}(x+3 b)=8 \quad[\quad x \neq-1] \\ & -1+3 b=8 \\ & 3 b=9 \\ \end{aligned}

b=3 \, \, \, \, \, \, \, \, \, ...(iii)

From the equations (ii) and (iii) the following is evident.

\begin{aligned} & a \\ & =-1-3 b \\ & =-1-3 \times 3 \\ & =-10 \\ & \therefore a<b \end{aligned}

 

Posted by

Ritika Kankaria

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