Question

For $f\left(x\right)=\frac{3x^2+ax+a+1}{x^2+x-2}$ and $g\left(x\right)=\frac{\log (1+x+x^2)+\log (1-x+x^2)}{\sec x-\cos x}$, which of the following are correct?

• A. $\underset{x\to 1}{lim}f\left(x\right)$ exists if $a=-2$
• B. $\underset{x\to 0}{lim}g\left(x\right)=1$
• C. If $\underset{x\to 1}{lim}f\left(x\right)$ exists and finite, then$\underset{x\to 1}{lim}f\left(x\right)=\frac{4}{3}$
• D. $\underset{x\to 0}{lim}g\left(x\right)=-1$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\underset{x\to 1}{lim}f\left(x\right)$ exists if $a=-2$
B $\underset{x\to 0}{lim}g\left(x\right)=1$
C If $\underset{x\to 1}{lim}f\left(x\right)$ exists and finite, then$\underset{x\to 1}{lim}f\left(x\right)=\frac{4}{3}$

For $f\left(x\right)=\frac{3x^2+ax+a+1}{x^2+x-2}$,
As $x\to 1$, denominator $\to 0$. Hence, Numerator $\to 0$
$\Rightarrow 3+a+a+1=0\Rightarrow a=-2$

Now, $\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}\frac{3x^2-2x-1}{x^2+x-2}$
$=\underset{x\to 1}{lim}\frac{(3x+1)(x-1)}{(x+2)(x-1)}$
$=\frac{4}{3}$

$\underset{x\to 0}{lim}g\left(x\right)=\underset{x\to 0}{lim}\frac{\log (1+x+x^2)+\log (1-x+x^2)}{\sec x-\cos x}$
$=\underset{x\to 0}{lim}\frac{\log \left[\right(1+x^2{)}^2-x^2]}{\frac{1-{\cos }^{2}x}{\cos x}}$
$=\underset{x\to 0}{lim}\frac{\log (1+x^2+x^4)}{\sin x\tan x}$
$=\underset{x\to 0}{lim}\frac{\log (1+x^2+x^4)}{x^2(1+x^2)}{x}^2(1+x^2)\frac{1}{\frac{\sin x}{x}\frac{\tan x}{x}{x}^2}$
$=1$