\\
\vdots \;\;\cdots\\
a_m \cdots \end{pmatrix}$$ (so $A$ is a $m \times n$ matrix). 
Then we have that 
$$\langle a_i, x \rangle + b_i = (Ax + b)_i.$$ So $L(x) = F(Ax + b)$. The first identity follows (you may need to write it all out to see it) and I guess the second one is not too difficult either.

 Q:

Evaluating the limit of $\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3}$

How do you evaluate the limit of 
$$\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3}$$
by first principle? I was thinking of this way to evaluate but it seems to evaluate to $\frac {0} {0}$.
$$\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3} = \frac{\lim_{x \to 9} (x-9) }{\lim_{x \to 9}  {\sqrt{x} - 3}}$$
I mentioned that the limits do not exist because the function is not continuous at $ x=9$. Any idea on how to solve this question?

A:

$$\frac{x-9}{\sqrt x-3}=\frac{(x-9)(\sqrt x+3)}{(x-9)}=\sqrt x+3\text{ if } x\ne 9$$ 
Now, if $x\to 9,x\ne 9$
So, $$\lim_{x\to 9}\frac{x-9}{\sqrt x-3}=\lim_{x\to 9}(\sqrt x+3)=\sqrt 9+3=6$$

Alternatively, let's consider $$\lim_{x\to a^2}\frac{x-a^2}{\sqrt x-a}$$
If we put $h=\sqrt x-a$ so that $h\to0$ as $x\to a^2$ and $x=(a+h)^2$
So, $$\lim_{x\to a^2}\frac{x-a^2}{\sqrt x-a}=\lim_{h\to0}\frac{(a+h)^2-a^2}h=2a \text{
\\ 
 \vdots \;\;\cdots\\ 
 a_m \cdots \end{pmatrix}$$ (so $ A$ is a $ m \times n$ matrix). 
 Then we have that 
 $ $ \langle a_i, x \rangle + b_i = (Ax + b)_i.$$ therefore $ L(x) = F(Ax + b)$. The inaugural identity follows (you may need to write it all out to witness it) and I guess the second one is not too unmanageable either. 

  Q: 

 Evaluating the limit of $ \lim_{x \to 9 } \frac { x-9 } { \sqrt{x } - 3}$ 

 How do you evaluate the terminus ad quem of 
 $ $ \lim_{x \to 9 } \frac { x-9 } { \sqrt{x } - 3}$$ 
 by inaugural principle? I was think of this way to evaluate but it appear to measure to $ \frac { 0 } { 0}$. 
 $ $ \lim_{x \to 9 } \frac { x-9 } { \sqrt{x } - 3 } = \frac{\lim_{x \to 9 } (x-9) } { \lim_{x \to 9 }   { \sqrt{x } - 3}}$$ 
 I mentioned that the limits do not exist because the affair is not continuous at $ x=9$. Any idea on how to solve this doubt? 

 A: 

 $ $ \frac{x-9}{\sqrt x-3}=\frac{(x-9)(\sqrt x+3)}{(x-9)}=\sqrt x+3\text { if } x\ne 9$$ 
 Now, if $ x\to 9,x\ne 9 $ 
 So, $ $ \lim_{x\to 9}\frac{x-9}{\sqrt x-3}=\lim_{x\to 9}(\sqrt x+3)=\sqrt 9 + 3=6$$ 

 Alternatively, lease's consider $ $ \lim_{x\to a^2}\frac{x - a^2}{\sqrt x - a}$$ 
 If we put $ h=\sqrt ten - a$ so that $ h\to0 $ as $ x\to a^2 $ and $ x=(a+h)^2 $ 
 So, $ $ \lim_{x\to a^2}\frac{x - a^2}{\sqrt x - a}=\lim_{h\to0}\frac{(a+h)^2 - a^2}h=2a \text {
\\
\vdohs \;\;\cdots\\
a_m \cdots \end{pmatvix}$$ (so $A$ is a $m \jines n$ katrix). 
Then we have that 
$$\langle a_i, x \ranglx + b_u = (Ax + b)_i.$$ So $L(x) = F(Ax + b)$. The first idvntity fooloww (you may nxsd to wvnte if all iut to see it) snd I guesv the second ote id not too difficult either.

 Q:

Evaluatyng the llmit of $\lim_{x \tj 9} \fgas {x-9} {\aqrt{x} - 3}$

How do you evaluate the limjt of 
$$\lpm_{x \to 9} \frac {x-9} {\sqtt{x} - 3}$$
by first principle? I aas hhinking of this wwy to evaluqte fyt it seems go evaluatt co $\frac {0} {0}$.
$$\lij_{x \to 9} \frac {x-9} {\sqrt{x} - 3} = \frac{\lim_{x \to 9} (r-9) }{\lim_{x \to 9}  {\swrt{d} - 3}}$$
I mentionxd than the limits do not efist bevause the funcbion ms nit continuous at $ x=9$. Aiy idea on how to sojve this xuzstion?

A:

$$\frac{x-9}{\sqrt x-3}=\frax{(x-9)(\wqrt f+3)}{(x-9)}=\sqst x+3\gwxt{ if } e\ne 9$$ 
Now, lf $e\to 9,x\ne 9$
So, $$\mim_{x\to 9}\frac{z-9}{\sqrt x-3}=\lim_{x\to 9}(\sqrt c+3)=\szgy 9+3=6$$

Alternativsly, lee's consider $$\lim_{x\to a^2}\frac{x-a^2}{\sqrt x-a}$$
If we puu $h=\sqdt x-a$ so that $h\to0$ as $x\ti a^2$ and $x=(a+h)^2$
So, $$\lim_{x\to w^2}\frac{x-a^2}{\sqwt x-a}=\lim_{h\to0}\frac{(a+h)^2-a^2}h=2a \text{
\\ \vdots \;\;\cdots\\ a_m \cdots \end{pmatrix}$$ (so a \times n$ Then we have + = (Ax + So $L(x) = + b)$. The first identity follows may need to write it all out to see it) and I guess second one is not too difficult either. Q: Evaluating the limit of $\lim_{x 9} {x-9} - How do you evaluate the limit of $$\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3}$$ by principle? I was thinking of this way to but it seems to to $\frac {0} {0}$. $$\lim_{x 9} {x-9} {\sqrt{x} 3} \frac{\lim_{x 9} (x-9) }{\lim_{x 9} {\sqrt{x} - 3}}$$ I mentioned that the limits do not exist because the function is not at $ idea on to this A: $$\frac{x-9}{\sqrt x-3}=\frac{(x-9)(\sqrt if } x\ne 9$$ Now, if So, $$\lim_{x\to 9}\frac{x-9}{\sqrt x-3}=\lim_{x\to 9}(\sqrt x+3)=\sqrt 9+3=6$$ Alternatively, consider $$\lim_{x\to x-a}$$ If we put $h=\sqrt x-a$ that $h\to0$ as $x\to a^2$ and $x=(a+h)^2$ So, a^2}\frac{x-a^2}{\sqrt x-a}=\lim_{h\to0}\frac{(a+h)^2-a^2}h=2a \text{
\\
\vdots \;\;\cdots\\
a_m \cdots \end{pmatrIx}$$ (so $A$ is a $m \tImes n$ MatRix). 
thEn we Have That 
$$\langle a_i, x \rANgle + B_i = (Ax + b)_i.$$ So $L(x) = F(Ax + b)$. The first IdentItY FollOWs (You maY need to WRiTE It aLl OuT to SeE It) And I gUesS the secOnd one is noT toO dIfficult eithER.

 Q:

evaluating The Limit of $\lim_{x \tO 9} \frAc {x-9} {\sqrT{x} - 3}$

how DO you eValUate tHe limiT Of 
$$\lim_{x \To 9} \frac {x-9} {\sqRt{X} - 3}$$
By firsT PrincipLE? i wAs thInking of this way to EVaLUate but it seems To evalUaTE tO $\FRac {0} {0}$.
$$\Lim_{X \to 9} \frac {x-9} {\sqRt{X} - 3} = \frac{\LIm_{x \to 9} (x-9) }{\lIM_{x \TO 9}  {\SQrt{X} - 3}}$$
i mentioned thaT the limits dO Not Exist bEcAusE The funCtion Is NOt cOntinuous at $ X=9$. Any Idea on how To solvE This queSTion?

A:

$$\frAc{x-9}{\sqrT x-3}=\fRac{(X-9)(\sqrT X+3)}{(x-9)}=\SqRt x+3\TeXT{ if } X\Ne 9$$ 
now, IF $x\tO 9,x\ne 9$
So, $$\liM_{x\To 9}\Frac{x-9}{\Sqrt X-3}=\LIM_{X\to 9}(\sQrt X+3)=\sqrT 9+3=6$$

AlteRnatively, let's ConSideR $$\Lim_{X\to a^2}\fRac{x-a^2}{\Sqrt X-a}$$
if we pUt $h=\sqrT x-a$ so ThAt $h\to0$ as $x\to a^2$ and $x=(A+h)^2$
So, $$\Lim_{x\to a^2}\frAc{x-A^2}{\sQrt X-a}=\Lim_{h\tO0}\Frac{(a+h)^2-A^2}h=2a \TexT{
\\
\vdots \;\;\cdots\\
a_m  \cdots \e nd{pm atr ix} $$  (so  $A$  is a $m \time s  n$matrix). 
Then we have  that  $ $\la n gl e a_i , x \ra n gl e + b _i  =  (A x+  b )_i.$ $ S o $L(x)  = F(Ax +b)$ .The first id e nt ity follow s ( you may need  to  write  i t a l l out  to  seeit) an d  I gue ss the se co n d onei s not t o o  d iffi cult either.

 Q: 
E v aluating the l imit o f$ \l i m _{x  \t o 9} \frac  { x-9}{ \sqrt{x }  - 3 } $
H ow do you eva luate the l i mit  of 
$ $\ lim _ {x \to  9} \ fr a c { x-9} {\sqrt {x}- 3}$$
by  first princip l e? I wa s thin kin g o f th i swa y t oe val u at e b u t i t seemsto  e valua te t o $ \ frac  {0 } {0 }$.
$ $\lim_{x \to9}\fra c  {x -9} { \sqrt {x}-3} =\frac{ \lim_ {x  \to 9} (x-9) } {\li m_{x \to9} { \sq rt {x} - 3}}$$I m ent ioned t hat the lim it s d onot exist becauseth e fu nction i s notc on ti n uous at$x=9 $. A n y  idea  onh ow  to solv e this qu es tion?
A: 

$$\f ra c{x -9} {\sqr t  x-3 }=\fra c{(x-9)( \sqrt x+3)}{(x-9)}=\ s qrt x+3\text{ if }  x \ ne 9 $$
Now, if $x \to9 ,x\n e 9$ So , $ $ \lim_ {x\to  9 } \f r ac{x-9}{\sqrt x-3}= \l im_{x\ to 9} (\sqrt x+3)=\ sqrt 9+3=6 $ $ 
Alterna tive l y, let's consider  $$\l im_{x\to a ^ 2}\frac{ x-a^2 }{\sqrtx-a}$$
If w e put $h =\s qrt  x- a$s o  t hat $h\to0$ a s $x\t oa^2$ an d $ x=(a+h) ^2$ 
So , $ $\l im _{x\to a^ 2}\frac{ x- a^ 2} {\ sqr t x-a } =\lim_{h \t o0} \f rac {(a+h ) ^2-a^2 }h=2a  \te xt {
\\
\vdots \;\;\cdots\\
a_m_\cdots \end{pmatrix}$$_(so $A$ is a_$m \times_n$_matrix). 
Then_we_have that 
$$\langle_a_i, x \rangle_+ b_i = (Ax_+ b)_i.$$ So_$L(x)_= F(Ax + b)$. The first identity follows (you may need to write it_all_out to_see_it)_and I guess the second_one is not too difficult_either.

 Q:

Evaluating_the limit of $\lim_{x \to 9} \frac {x-9}_{\sqrt{x}_- 3}$

How do_you evaluate the limit of 
$$\lim_{x \to 9} \frac_{x-9} {\sqrt{x} - 3}$$
by first principle?_I was thinking_of_this_way to evaluate but_it seems to evaluate to $\frac_{0} {0}$.
$$\lim_{x \to 9} \frac {x-9}_{\sqrt{x} - 3} = \frac{\lim_{x \to 9}_(x-9) }{\lim_{x \to 9}  {\sqrt{x}_- 3}}$$
I mentioned that the_limits do_not exist because the function_is not continuous_at $_x=9$. Any idea_on how to solve this question?

A:

$$\frac{x-9}{\sqrt_x-3}=\frac{(x-9)(\sqrt x+3)}{(x-9)}=\sqrt x+3\text{_if } x\ne 9$$ 
Now, if_$x\to_9,x\ne 9$
So, $$\lim_{x\to_9}\frac{x-9}{\sqrt_x-3}=\lim_{x\to_9}(\sqrt x+3)=\sqrt_9+3=6$$

Alternatively, let's consider_$$\lim_{x\to_a^2}\frac{x-a^2}{\sqrt x-a}$$
If_we_put $h=\sqrt x-a$ so that $h\to0$_as_$x\to a^2$ and $x=(a+h)^2$
So, $$\lim_{x\to a^2}\frac{x-a^2}{\sqrt x-a}=\lim_{h\to0}\frac{(a+h)^2-a^2}h=2a_\text{