Solution:
A function f(x) is said to be continuous if the limit $\lim_{x\to a} f(x)$ exists and it is matching the actual value of the function irrespective of in what direction (from left or from right) we approach the point along the function. \[\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = f(a)\]

Correct Answer: D

Solution:

S = {x :x \(\in\) A AND x \(\in\) B} represents the INTERSECTION of set A and set B.
S = {x :x \(\in\) A OR x \(\in\) B} represents the UNION of set A and set B.

Correct Answer: C

Solution:

For x > 0, f'(x)=1, $\lim_{x\to 0^+}$ f'(x)=1

For x < 0, f'(x)=-1, $\lim_{x\to 0^-}$ f'(x)= -1

$\lim_{x\to 0^+}$ f'(x) $\neq$ $\lim_{x\to 0^-}$ f'(x)

Correct Answer: C

Solution:
\[y=\lim _{x\to 0}\dfrac {\left( 1-\cos x\right) }{x^{2}}\] It forms 0/0 condition. Hence by L'Hopital rule, \[y=\lim _{x\to 0}\dfrac {\dfrac {d}{dx}\left( 1-\cos x\right) }{\dfrac {d}{dx}x^{2}}\] Still this gives $\dfrac{0}{0}$ condition, so again applying L'Hopital rule, \[y =\lim _{x\to 0}\dfrac {\dfrac {d}{dx}\left( \sin x \right) }{\dfrac {d}{dx}\left( 2x\right)} =\lim _{x\to 0}\dfrac {\cos x}{2} =\dfrac{\cos 0}{2} = \dfrac{1}{2}\]

Correct Answer: B

Solution:
This is a standard identity or you may apply L'Hopital's rule, \[y =\lim_{\theta \to 0}\dfrac {\sin \theta }{\theta } =\lim_{\theta \to 0}\dfrac {\dfrac {d}{d\theta }\sin \theta }{\dfrac {d}{d\theta }\theta } =\lim_{\theta \to 0}\dfrac {\cos \theta }{1} =\dfrac {\cos 0}{1} =1\]

Correct Answer: D

Solution:
\begin{align*} f\left( x\right) &=\lim _{x\rightarrow 0}\dfrac {\sin ^{2}x}{x}\\ &=\lim _{x\rightarrow 0}\dfrac {\sin ^{2}x}{x^{2}}\times \dfrac {x^{2}}{x}\\ &=\lim _{x\rightarrow 0}\left( \dfrac {\sin x}{x}\right) ^{2}\times x\\ &=\left( 1\right) ^{2}\times 0\\ &=0 \end{align*} Alternative method is to use the L'Hospital's Rule. Take derivative of both numerator as well as denominator. \begin{align*} f\left( x\right) &=\lim _{x\rightarrow 0}\dfrac {\sin ^{2}x}{x}f\left( x\right)\\ &=\lim _{x\rightarrow 0}\dfrac {2\sin x\cos x}{1}\\ &=\lim _{x\rightarrow 0}\dfrac {\sin 2x}{1}=\dfrac {\sin 0}{1}\\ &=0 \end{align*}

Correct Answer: A

Solution:

Correct Answer: D

Solution:

Correct Answer: D

Solution:
Given :
y = x$^2$ ...(i) and interval [1, 5]
At x = 1 and y = 1,
And at x = 5, y = (5)$^2$ = 25
Here the interval is bounded between 1 and 5. So, the minimum value at this interval is 1.

Correct Answer: B

Solution:
Using the formula for, $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
Here, a = x$^{\frac{1}{3}}$
and b = 2 \begin{align*} \lim _{x\rightarrow 8}\frac {x^{\frac{1}{3}}-2}{x-8} &= \dfrac {(x^{\frac{1}{3}}-2)}{(x^{\frac{1}{3}}-2)(x^{\frac{2}{3}} + {x^{\frac{1}{3}}(2) +4)}}\\ &= \dfrac {1}{(4)+ 2(2)+4}\\ &= \dfrac {1}{12} \end{align*}

Correct Answer: B

Solution:
First we will factorize numerator and denominator separately. \begin{align*} 2x^2 -7x +3 &=2x^{2}-6x-x+3\\ &=2x\left( x-3\right) -1\left( x-3\right)\\ &=\left( x-3\right) \left( 2x-1\right)\\ \end{align*} \begin{align*} 5x^{2}-12x-9 &=5x^{2}-15x+3x-9\\ &=5x\left( x-3\right) +3\left( x-3\right)\\ &=\left( x-3\right) \left( 5x+3\right) \end{align*} After cancelling the common factor from numerator and denominator, $f\left( x\right) =\dfrac {2x-1}{5x+3}$
Now substitute the value of x = 3, $ f(x)=\dfrac {5}{18}$

Correct Answer: B

Solution:
\[\mathrm{Let\ } L = \lim_{x \to 1} \left( 1-x\right) \tan \left( \dfrac {\pi x}{2}\right)\] \[\mathrm{Put}\ 1-x = \theta, \quad x=1-\theta, \quad \mathrm{As\ } x \rightarrow 1,\ \theta \rightarrow 0\] \begin{align*} L &= \lim_{\theta \to 0} \theta \tan \left[\dfrac {\pi }{2}\left( 1-\theta \right)\right] \\ &= \lim_{\theta \to 0}\theta \tan \left( \dfrac {\pi }{2}-\dfrac {\pi }{2}\theta \right)\\ &= \lim_{\theta \to 0}\theta \left[ \cot \left( \dfrac {\pi }{2}\theta \right) \right]\\ &= \lim_{\theta \to 0}\dfrac {\theta }{\tan \left( \dfrac {\pi }{2}\theta \right) }\\ &=\lim _{\theta \rightarrow 0}\dfrac {{\theta }}{\left[ \dfrac {\tan \left( \dfrac {\pi }{2}\theta \right) }{\left(\dfrac {\pi }{2}\theta \right) }\right]\cdot \dfrac {\pi }{2}{\theta }} =\dfrac{1}{1 \cdot \left(\dfrac {\pi }{2}\right)} =\dfrac {2}{\pi} \end{align*}

Correct Answer: C

Solution:
\[f\left( h\right) =\left( 3h^{2}+4\right) \cos\ h,\] \[f\left( -h\right) =\left[ 3\left( -h\right) ^{2}+4\right] \cos \left( -h\right) =\left( 3h^{2}+4\right) \cos\ h\] Adding, $f\left( h\right) +f\left( -h\right) =\left( 6h^{2}+8\right) \cos\ h$
Subtract 8 from both sides, $f\left( h\right) +f\left( -h\right) -8 =6h^{2}\cos\ h +8\cos\ h -8$ \begin{align*} & \lim_{h \to 0} \dfrac {f\left( h\right) +f\left( -h\right) -8}{h^{2}}\\ &= \lim_{h \to 0} \dfrac{6h^{2}\cos\ h +8\cos\ h -8}{h^2}\\ &= \lim_{h \to 0} \dfrac {6h^{2}\cos\ h }{h^{2}}+8 \left(\dfrac {\cos\ h -1}{h^{2}}\right)\\ &= \lim_{h \to 0}6\cos\ h + 8\dfrac {\left(- 2\sin ^{2}\dfrac {h}{2}\right) }{h^{2}}\\ &=\lim_{h \to 0} 6\cos\ h - 16\left[ \dfrac{\left(\sin ^{2}\dfrac {h}{2}\right)}{\left( \dfrac {h}{2}\right) ^{2} \times 4}\right]\\ &=\lim_{h \to 0} 6\cos\ h -4 \lim_{\dfrac{h}{2} \to 0} \left[\dfrac {\sin ^{2}\left( \dfrac {h}{2}\right) }{\left( \dfrac {h}{2}\right) ^{2}}\right]\\ &=6(1)-4(1) =2 \end{align*}

Correct Answer: B

Solution:
\[\mathrm{Let\ L =} \lim_{x \to \infty} \dfrac{x \ln x}{1 + x^2}\] Applying L'Hopital's rule and taking derivatives of both numerator and denominator: \[= \lim_{x \to \infty} \left[\dfrac{x\left(\dfrac{1}{x}\right) + \ln x}{2x} \right]\] Still this limit is in $\dfrac{\infty}{\infty}$ form, hence applying L'Hopital's rule one more time \begin{align*} L &= \lim_{x \to \infty} \left[\dfrac{0 + \dfrac{1}{x}}{2} \right]\\ &= \lim_{x \to \infty} \left[\dfrac{1}{2x} \right]\\ &= \dfrac{1}{2 \times \infty}\\ &= 0 \end{align*}

Correct Answer: D