Question

The limit with the steps if it exist plz

Answer 1

Answer:

1/4

Step-by-step explanation:

Hello, first of all we could say, let 's replace x by 1 and see if we can conclude.

numerator gives $\sqrt{1+15}-4=\sqrt{16}-4=4-4=0$

denominator gives 1-1=0

So, this is 0/0 and this is not defined.

We need to ask a friend for help. Guillaume de l'Hôpital, French mathematician from the 1600s,  has a trick to solve this kind of stuff.

In short, he says that

$\displaystyle \lim_{x\rightarrow c} \ {\dfrac{f(x)}{g(x)}}=\lim_{x\rightarrow c} \ {\dfrac{f'(x)}{g'(x)}}$

In our case here, we have c = 1

$f(x)=\sqrt{x^2+15}-4\\\\g(x)=x-1$

$f'(x)=\dfrac{1}{2}\dfrac{2x}{\sqrt{x^2+15}}=\dfrac{x}{\sqrt{x^2+15}}\\\\f'(1)=\dfrac{1}{4}\\\\g'(x)=1\\\\\dfrac{f'(1)}{g'(1)}=\dfrac{1}{4}$

So, we can conclude

$\displaystyle \lim_{x\rightarrow1} \ {\dfrac{\sqrt{x^2+15}-4}{x-1}}=\lim_{x\rightarrow1} \ {\dfrac{1}{4}}=\boxed{\dfrac{1}{4}}$

Thanks