0/0 and ∞/∞ Form — JEE Main Maths PYQ MCQs with Solutions

Solution: Both numerator and denominator vanish at $x=0$ (0/0 form). By L'Hopital's rule, differentiate numerator and denominator: numerator derivative at $0$ is $2\cos 0-1=1$; denominator derivative is $\dfrac{x+\cos x}{\sqrt{x^2+2\sin x+1}}-\dfrac{\sin2x+1}{2\sqrt{\sin^2x+x+1}}$, which at $x=0$ equals $1-\dfrac12=\dfrac12$. So the limit $=\dfrac{1}{1/2}=2$.

Solution: For the limit to be finite, the numerator must vanish at $x=1$: $1-a+b=0\Rightarrow b=a-1$. By L'Hopital, $\lim_{x\to1}(2x-a)=2-a=5\Rightarrow a=-3$. Then $b=a-1=-4$, so $a+b=-3-4=-7$.

Solution: LHS $=\lim_{x\to1}(x^3+x^2+x+1)=4$. RHS: $\dfrac{x^3-k^3}{x^2-k^2}=\dfrac{x^2+xk+k^2}{x+k}\to\dfrac{3k^2}{2k}=\dfrac{3k}{2}$ as $x\to k$. Setting $\dfrac{3k}{2}=4$ gives $k=\dfrac{8}{3}$.

Solution: Rationalise: multiply by $\sqrt2+\sqrt{1+\cos x}$. Denominator becomes $2-(1+\cos x)=1-\cos x$. Since $\sin^2x=(1-\cos x)(1+\cos x)$, the expression simplifies to $(1+\cos x)(\sqrt2+\sqrt{1+\cos x})$, which at $x=0$ gives $2(\sqrt2+\sqrt2)=4\sqrt2$.

Solution: $\cot^3x-\tan x=\dfrac{\cos^4x-\sin^4x}{\sin^3x\cos x}=\dfrac{\cos2x}{\sin^3x\cos x}$. Also $\cos\left(x+\frac{\pi}{4}\right)=\dfrac{\cos x-\sin x}{\sqrt2}$ and $\cos2x=(\cos x-\sin x)(\cos x+\sin x)$. Cancelling $(\cos x-\sin x)$ gives $\dfrac{\sqrt2(\cos x+\sin x)}{\sin^3x\cos x}$, which at $x=\pi/4$ (where $\sin x=\cos x=1/\sqrt2$) equals $\dfrac{\sqrt2\cdot\sqrt2}{1/4}=8$.

Solution: Using $\cot\theta\sim\dfrac{1}{\theta}$ and $\sin x\sim x$ for small $x$: numerator $\sim x\cdot\dfrac{1}{4x}=\dfrac14$; denominator $\sim x^2\cdot\dfrac{1}{4x^2}=\dfrac14$. Ratio $\to 1$.

Solution: Let $t=y^4\to0^+$. $\sqrt{1+t}\approx1+t/2$, so $1+\sqrt{1+t}\approx2+t/2=2\left(1+\dfrac{t}{4}\right)$, giving $\sqrt{1+\sqrt{1+t}}\approx\sqrt2\left(1+\dfrac{t}{8}\right)$. So the expression $\approx\sqrt2\cdot\dfrac{t}{8t}=\dfrac{1}{4\sqrt2}$, and the limit exists.

Solution: Put $x=\pi/2-h$, $h\to0$. Then $\cot x=\tan h\approx h+h^3/3$, $\cos x=\sin h\approx h-h^3/6$, so $\cot x-\cos x\approx h^3/2$. Also $\pi-2x=2h$, so $(\pi-2x)^3=8h^3$. The limit $=\dfrac{h^3/2}{8h^3}=\dfrac{1}{16}$.

Solution: $\cos^2x=1-\sin^2x$, so $\pi\cos^2x=\pi-\pi\sin^2x$ and $\sin(\pi\cos^2x)=\sin(\pi\sin^2x)\approx\pi\sin^2x\approx\pi x^2$ for small $x$. So the limit is $\pi$.

Solution: $1-\cos2x=2\sin^2x\approx2x^2$, $(3+\cos x)\to4$, $\tan4x\approx4x$. So the expression $\approx\dfrac{2x^2\cdot4}{x\cdot4x}=2$.

Solution: $\dfrac{x^2+x+1}{x+1}=x+\dfrac{1}{x+1}$. For the limit as $x\to\infty$ to be finite we need $a=1$; then the expression becomes $\dfrac{1}{x+1}-b\to -b=4\Rightarrow b=-4$.

Solution: As $h\to0$, $2h+2+h^2\to2$ with rate $2+2h\to2$, so the numerator $\sim f'(2)\cdot2h$. Similarly $h-h^2+1\to1$ with rate $1-2h\to1$, so denominator $\sim f'(1)\cdot h$. Ratio $\to\dfrac{f'(2)\cdot2}{f'(1)}=\dfrac{6\cdot2}{4}=3$.

Solution: Since $\sin nx\approx nx$, the expression $\approx\dfrac{n[(a-n)nx-\tan x]}{x}$. Using $\tan x\approx x+x^3/3$, this tends to $n[(a-n)n-1]$. Setting this equal to $0$ gives $(a-n)n=1\Rightarrow a=n+\dfrac1n$.

Solution: $\cos x-1\approx-\dfrac{x^2}{2}$ and $\cos x-e^x\approx(1-\tfrac{x^2}{2})-(1+x+\tfrac{x^2}{2})=-x-x^2\approx-x$. Their product $\approx\dfrac{x^3}{2}$, so $n=3$ makes the limit finite and non-zero.

Solution: Using $\tan\theta\approx\theta+\theta^3/3$: $x\tan2x\approx2x^2+\tfrac{8x^4}{3}$ and $2x\tan x\approx2x^2+\tfrac{2x^4}{3}$, so the numerator $\approx2x^4$. Also $1-\cos2x=2\sin^2x\approx2x^2$, so the denominator $\approx4x^4$. The limit is $\dfrac{2x^4}{4x^4}=\dfrac12$.

Solution: Let $t=x-1$. $\sqrt{1-\cos2t}=\sqrt{2\sin^2t}=\sqrt2|\sin t|$. As $t\to0^+$, $\dfrac{\sqrt2\sin t}{t}\to\sqrt2$; as $t\to0^-$, $\dfrac{-\sqrt2\sin t}{t}\to-\sqrt2$. Since the left and right hand limits differ, the limit does not exist.

Solution: $\sqrt{\tfrac12(1-\cos^2x)}=\sqrt{\tfrac12\sin^2x}=\dfrac{|\sin x|}{\sqrt2}$. As $x\to0^+$, the expression $\to\dfrac{1}{\sqrt2}$; as $x\to0^-$, it $\to-\dfrac{1}{\sqrt2}$. The two-sided limit does not exist, so the answer is None of these.

Solution: For $x\in(-1,0)$, $[x]=-1$, so $f(x)=\dfrac{\sin(-1)}{-1}=\sin1$, giving left hand limit $\sin1$. For $x\in[0,1)$, $[x]=0$, so $f(x)=0$, giving right hand limit $0$. Since $\sin1\ne0$, the limit does not exist — answer is None of these.

Solution: The sum is $\dfrac{1+2+\cdots+n}{1+n^2}=\dfrac{n(n+1)/2}{1+n^2}$. Dividing numerator and denominator by $n^2$ and letting $n\to\infty$ gives $\dfrac12$.

Solution: By L'Hopital's rule, the limit equals $g'(a)f(a)-g(a)f'(a)=2\cdot2-(-1)\cdot1=4+1=5$.

Solution: This limit is simply $G'(1)$. Since $G'(x)=\dfrac{-x}{\sqrt{25-x^2}}$, we get $G'(1)=\dfrac{-1}{\sqrt{24}}$.

Solution: Each term telescopes: $\tan^{-1}\dfrac{1}{1+(x+j)(x+j-1)}=\tan^{-1}(x+j)-\tan^{-1}(x+j-1)$, so $f_n(x)=\tan^{-1}(x+n)-\tan^{-1}x$. At $x=0$, $f_j(0)=\tan^{-1}j$, so $\tan(f_j(0))=j$; hence $\sum_{j=1}^5j^2=55$, proving (a). Also $f_j'(0)=\dfrac{1}{1+j^2}-1$, so $(1+f_j'(0))\sec^2(f_j(0))=\dfrac{1}{1+j^2}(1+j^2)=1$, and summing $j=1$ to $10$ gives $10$, proving (b). As $x\to\infty$, $f_n(x)\to0$, so $\tan(f_n(x))\to0$ (not $1/n$, so (c) is false) and $\sec^2(f_n(x))\to\sec^2(0)=1$, proving (d).

Solution: Expand $\sqrt{a^2-x^2}=a-\dfrac{x^2}{2a}-\dfrac{x^4}{8a^3}-\cdots$, so $a-\sqrt{a^2-x^2}-\dfrac{x^2}{4}=\left(\dfrac{1}{2a}-\dfrac14\right)x^2+\dfrac{x^4}{8a^3}+\cdots$. For $L$ to be finite (dividing by $x^4$), the $x^2$ coefficient must vanish: $\dfrac{1}{2a}=\dfrac14\Rightarrow a=2$. Then $L=\dfrac{1}{8a^3}=\dfrac{1}{64}$.

Solution: $\sin(\beta x)\approx\beta x$, so the numerator $\approx\beta x^3$. The denominator $\alpha x-\sin x\approx(\alpha-1)x+\dfrac{x^3}{6}$. For the limit to be finite and non-zero, we need $\alpha=1$, leaving denominator $\approx x^3/6$. Then the limit becomes $\dfrac{\beta x^3}{x^3/6}=6\beta=1\Rightarrow\beta=\dfrac16$. So $6(\alpha+\beta)=6\left(1+\dfrac16\right)=7$.

Solution: $\cos(\alpha^n)\approx1-\dfrac{\alpha^{2n}}{2}$, so $e^{\cos(\alpha^n)}\approx e\cdot e^{-\alpha^{2n}/2}\approx e\left(1-\dfrac{\alpha^{2n}}{2}\right)$, giving $e^{\cos(\alpha^n)}-e\approx-\dfrac{e\alpha^{2n}}{2}$. Dividing by $\alpha^m$ gives $-\dfrac{e}{2}\alpha^{2n-m}$, which is finite and equal to $-\dfrac{e}{2}$ only if $2n-m=0$, i.e. $m=2n$. So $\dfrac{m}{n}=2$.