Question

If $f\left(x\right)=\underset{p\to \infty }{lim}\frac{x^{p}g\left(x\right)+h\left(x\right)+7}{7x^p+3x+1};x\ne 1$ and $f\left(1\right)=7,f\left(x\right),g\left(x\right)$ and $h\left(x\right)$ are all continuous functions at $x=1$ . Then which of the following statement(s) is/are correct

• A. $g\left(1\right)+h\left(1\right)=70$
• B. $g\left(1\right)-h\left(1\right)=28$
• C. $g\left(1\right)+h\left(1\right)=60$
• D. $g\left(1\right)-h\left(1\right)=-28$

Answer 1

Answer: (A), (B)

$g\left(1\right)+h\left(1\right)=70$
$g\left(1\right)-h\left(1\right)=28$

$f\left(x\right)={lim}_{p\to \infty }\frac{x^{p}g\left(x\right)+h\left(x\right)+7}{7x^p+3x+1}$

For $x\to 1^{+}$, Diving by $x^p$in the numerator and the denominator and $p\to \infty$ we get $f\left(1^{+}\right)=\frac{g\left(1\right)}{7}$

$\Rightarrow g\left(1\right)=7f\left(1\right)=49\to \left(1\right)$

Also $x=1f\left(1\right)=\frac{g\left(1\right)+h\left(1\right)+7}{11}$

$\Rightarrow f\left(1^{+}\right)=\frac{g\left(1\right)+h\left(1\right)+7}{11}=f\left(1^{-}\right)=7$

$f,g,h$ are continuous functions

$\Rightarrow g\left(1\right)+h\left(1\right)=70$

$g\left(1\right)=49\Rightarrow h\left(1\right)=21\Rightarrow g\left(1\right)-h\left(1\right)=28$