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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
lim⁡x→∞xp+xp−1+1xq+xq−2+2\displaystyle\lim_{x\to\infty}\dfrac{x^{p}+x^{p-1}+1}{x^{q}+x^{q-2}+2} (p>0,q>0p>0,q>0)
A00 if p<qp<qcorrect
B11 if p=qp=qcorrect
CInfinite if p>qp>qcorrect
D11 if p>qp>q
Solution
Step 1: Identify dominant powers. As x→∞x\to\infty, the numerator behaves like xpx^{p} and the denominator like xqx^{q}. Step 2: Divide numerator and denominator by xpx^{p}:
xp+xp−1+1xq+xq−2+2=1+1x+1xpxq−p+xq−p−2+2xp\dfrac{x^{p}+x^{p-1}+1}{x^{q}+x^{q-2}+2}=\dfrac{1+\dfrac{1}{x}+\dfrac{1}{x^{p}}}{x^{q-p}+x^{q-p-2}+\dfrac{2}{x^{p}}}
 Step 3: As x→∞x\to\infty, the numerator →1+0+0=1\to 1+0+0=1. Step 4: Case p<qp<q: Then q−p>0q-p>0, so xq−p→∞x^{q-p}\to\infty. Limit =1∞=0=\dfrac{1}{\infty}=0. Option (1) correct. Step 5: Case p=qp=q: Then xq−p=1x^{q-p}=1 and xq−p−2=x−2→0x^{q-p-2}=x^{-2}\to 0. Limit =11+0+0=1=\dfrac{1}{1+0+0}=1. Option (B) correct. Step 6: Case p>qp>q: Then q−p<0q-p<0, so xq−p→0x^{q-p}\to 0. Limit =10=∞=\dfrac{1}{0}=\infty. Option (3) correct, (4) wrong. Correct answers: (1), (2) and (3)
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