Question

${}_{x\to \infty }^{lim}\frac{x^p+x^{p-1}+1}{x^q+x^{q-2}+2}$, where $p>0,q>0$ is

• A. $0$ if $p<q$
• B. $1$ if $p=q$
• C. $\infty$, if $p>q$
• D. $1$ if $p<q$

Answer 1

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

The correct options are
A $0$ if $p<q$
B $1$ if $p=q$
C $\infty$, if $p>q$

Divide by $x^q$ and Apply exponent rule : $\frac{x^a}{x^b}=x^{a-b}$

$={lim}_{x\to \infty }\frac{x^{p-q}+x^{p-q-1}+\frac{1}{x^q}}{1+\frac{1}{x^2}+\frac{2}{x^q}}$

Put the value $x=\infty$

$={lim}_{x\to \infty }\frac{x^{p-q}+x^{p-q-1}+\frac{1}{\infty }}{1+\frac{1}{\infty }+\frac{2}{\infty }}$

$={lim}_{x\to \infty }(x^{p-q}+x^{p-q-1})$

Case I: If $p<q$, $(p=2,q=4)$

$={lim}_{x\to \infty }(x^{-2}+x^{-3})$

$={lim}_{x\to \infty }(\frac{1}{x^2}+\frac{1}{x^3})$

$=\frac{1}{\infty }+\frac{1}{\infty }$

$=0$

Option A is Correct

Case II : If $p=q$, $(p=q=1)$

$={lim}_{x\to \infty }(x^0+x^{-1})$

Simplification:

${lim}_{x\to \infty }\left(x^0\right)={lim}_{x\to \infty }\left(1\right)$

${lim}_{x\to a^c}=c=1$

$=1+{lim}_{x\to \infty }\left(\frac{1}{x}\right)$

$=1+\frac{1}{\infty }$

$=1$

Option B is Correct

Case III: If $p>q$, $(p=4,q=2)$

$={lim}_{x\to \infty }(x^2+x)$

$=\infty +\infty$

$=\infty$

Option C is Correct

Case IV: If $p<q$

Similar to Case I : (If p=2, q=4)

$={lim}_{x\to \infty }(x^{p-q}+x^{p-q-1})=0$

Option D is not Correct