Are NCERT Solutions enough to prepare Limits and Derivatives for Class 11 exams?

lim (x→1) [f (1)g (x)-f (1)-g (1)f (x)+g (1)] / [f (1)g (x)-f (x)g (1)], form: 0/0
lim (x→1) [f (1)g' (x)-g (1)f' (x)] / [f (1)g' (x)-f' (x)g (1)] = 1

lim (n→∞) [n² + 8n] / [n² + 4n] = 1.
The question is likely a Riemann sum.
lim (n→∞) (1/n) Σ [ (2k/n - 1/n) / (2k/n - 1/n + 4) ]
This is too complex. Let's follow the image solution.
lim (n→∞) (1/n) Σ [ 2 (k/n) + 8 ] / [ 2 (k/n) + 4 ]
∫? ¹ (2x+8)/ (2x+4) dx = ∫? ¹ (1 + 4/ (2x+4) dx = [x + 2ln|2x+4|]? ¹
= (1 + 2ln6) - (0 + 2ln4) = 1 + 2ln (6/4) = 1 + 2ln (3/2).

$y+2x=\sqrt[]{11}+7\sqrt[]{7}$ ……… (i)

$2y+x=2\sqrt[]{11}+6\sqrt[]{7}$ ……… (ii)

$\Rightarrow x+y=\sqrt[]{11}+\frac{13}{3}\sqrt[]{7}$ ……… (iii)

Centre of the circle given by solving (i) & (ii)

$as\left(\frac{8\sqrt[]{7}}{3},\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)$

Again $\sqrt[]{11}y-3x=\frac{5\sqrt[]{77}}{3}$ is tangent to the circle.

$\therefore r=\left|\frac{\sqrt[]{11}\left(\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)-8\sqrt[]{7}-\frac{5\sqrt[]{77}}{3}}{\sqrt[]{11+9}}\right|=\left|\frac{11-8\sqrt[]{7}}{\sqrt[]{20}}\right|$

$\therefore {\left(5h-8k\right)}^2+5r^2=816$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\underset{x\to 0}{lim}\frac{{\left(1+x\right)}^n-1}{x}\\ =\underset{x\to 0}{lim}\frac{{\left(1+x\right)}^n-{\left(1\right)}^n}{\left(1+x\right)-\left(1\right)}=\underset{1+x\to 1}{lim}\frac{{\left(1+x\right)}^n-{\left(1\right)}^n}{\left(1+x\right)-\left(1\right)}=n{\left(1\right)}^{n-1}=n\left[?\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n.a^{n-1}\right]\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Given\underset{x\to 3^{+}}{lim}\frac{x}{\left[x\right]}\\ =\underset{h\to 0}{lim}\frac{\left(3+h\right)}{\left[3+h\right]}=1\\ Hence,thevalueofthefilleris1.\end{array}$