MCQ
What is the value $ \lim_{\text{x} \rightarrow 4}\frac{\text{x}^2-2\text{x}-8}{\text{x}-4}$ :
- A0
- B2
- C8
- D6
Solution:
The denominator becomes 0, as x approaches 4.
$ \lim_{\text{x} \rightarrow 4}\frac{\text{x}^2-2\text{x}-8}{\text{x}-4}$ Here, if we factorize the numerator we get
$ \lim_{\text{x} \rightarrow 4}\frac{(\text{x}-4) (\text{x}+2)}{\text{x}-4}$
We can now cancel out (x - 4) from both the numerator and denominator.
We get, $ \lim_{\text{x} \rightarrow 4}(\text{x}+2)=6$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
If $\text{f}(\text{x})=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$ then $\frac{\text{dy}}{\text{dx}}$ at x = 1 is equal to:
$1$
$\frac{1}{2}$
$\frac{1}{\sqrt{2}}$
$0$
$\text{A}:\cos\text{a}+\cos\text{b}+\cos\text{g}=0$
$\text{B}:\sin\text{a}+\sin\text{b}+\sin\text{g}=0$
If
$\cos(\beta-\text{y})+\cos(\text{y}-\alpha)+\cos(\alpha-\beta)=\frac{-3}{2}$ then: