Question

Hello Teacher, Could you help me with this equation?​

Answer 1

Answer:

$\frac{1}{42}$

Step-by-step explanation:

Since $\frac{\sqrt{0+9}-3 }{sin7n} = \frac{0}{0}$

We should apply the L'Hopital Rule to find the limit. (Use this when we see $\frac{0}{0}$.

Take the derivative of the numerator and the denominator individually.

$\frac{d}{dn} (\sqrt{n+9}-3) =$ $\frac{1}{2\sqrt{n+9} }$

$\frac{d}{dn}(sin7n)$ $= 7cos(7n)$

So the new equation is:

$\frac{1}{14cos(7n)\sqrt{n+9}}$

Now we can plug in n = 0 into the new equation:

$\frac{1}{14cos(7*0)\sqrt{0+9}}$

Since $cos(7*0) = cos(0) = 1$

$\frac{1}{14*1\sqrt{9}}$

$\frac{1}{14*3} = \frac{1}{42}$

Therefore the answer is $\frac{1}{42}$.