Substitution Method to Remove Indeterminate Form

Q.

$\underset{n\to \infty }{lim}\sum _{r=1}^{n}co{t}^{-1}(r^2+\frac{3}{4})$ is

1. 0

2. $ta{n}^{-1}1$

3. $ta{n}^{-1}2$

4. ${\tan }^{-1}\frac{1}{2}$

Q. A right angled triangle has one leg of length $1$ unit, another leg of length $x$ units and hypotenuse is of length $y$ units. The angle opposite to the side of length $x$ units is $\theta .$ Then

1. $y^3–x^3={\sec }^3\theta –{\tan }^3\theta$
2. $\underset{\theta \to \pi /2}{lim}(y-x)=0$
3. $\underset{\theta \to \pi /2}{lim}(y^2-x^2)=1$
4. $\underset{\theta \to \pi /2}{lim}(y^3-x^3)=-1$

Q. The value of $\underset{n\to \infty }{\lim }\frac{n!}{(n+1)!-n!}$ is

Q. Let $M$ and $m$ respectively be the maximum and minimum values of the function $f\left(x\right)={\tan }^{-1}(\sin x+\cos x)\text{in}[0,\frac{\pi }{2}].$ Then the value of $\tan (M-m)$ is equal to

1. $2-\sqrt{3}$
2. $2+\sqrt{3}$
3. $3+2\sqrt{2}$
4. $3-2\sqrt{2}$

Q. If $f\left(x\right)=\{\begin{array}{cc}\frac{(1-\sin x)}{(\pi -2x{)}^2},& x\ne \frac{\pi }{2}\\ \lambda ,& x=\frac{\pi }{2}\end{array}$ is continuous at $x=\frac{\pi }{2}$, then the value of $\lambda$ is

1. $\frac{1}{8}$
2. $\frac{1}{2}$
3. $\frac{1}{4}$
4. $1$

Q.

Find the value of ${lim}_{x\to 3}\frac{x^2-8x+15}{x^2-9x+18}$

1. $\frac{2}{3}$

2. $\frac{3}{2}$

3. $\frac{4}{3}$

4. $\frac{1}{2}$

Q. The value of $\underset{x\to 1}{lim}\frac{1-\sqrt{x}}{({\cos }^{-1}x{)}^2}$ is

1. $4$
2. $\frac{1}{2}$
3. $2$
4. $\frac{1}{4}$