Question

Giúp mik với ạ

Answer 1

\(a=\lim\limits_{x\rightarrow-3}\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{1}{x-3}=-\dfrac{1}{6}\)

\(b=\lim\limits_{x\rightarrow2}\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\lim\limits_{x\rightarrow2}\dfrac{x+3}{x+2}=\dfrac{5}{4}\)

\(c=\lim\limits_{x\rightarrow4}\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x+5\right)\left(x-4\right)}=\lim\limits_{x\rightarrow4}\dfrac{x+4}{x+5}=\dfrac{8}{9}\)

\(d=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\dfrac{x+2}{x-1}=4\)

\(e=\lim\limits_{x\rightarrow2}\dfrac{x+7-9}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{1}{\sqrt{x+7}+3}=\dfrac{1}{6}\)

\(f=\lim\limits_{x\rightarrow1}\dfrac{x+3-4}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}\)

\(h=\lim\limits_{x\rightarrow-3}\dfrac{x+7-4}{\left(x+3\right)\left(\sqrt{x+7}+2\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+3}{\left(x+3\right)\left(\sqrt{x+7}+2\right)}=\lim\limits_{x\rightarrow-3}\dfrac{1}{\sqrt{x+7}+2}=\dfrac{1}{4}\)

Answer 2

Bài 1:

a, 

= lim_x->-3 \(\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}\)

= lim_x->3  x-3

= -3 -3

= -6

b, 

= lim_x->2 \(\dfrac{\left(x-2\right)\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}\)

= lim_x->2  \(\dfrac{x+3}{x+2}\)

= \(\dfrac{5}{4}\)

c,

= lim_x->4   \(\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+5\right)}\)

= lim_x->4   \(\dfrac{\left(x+4\right)}{\left(x+5\right)}\)

= \(\dfrac{8}{9}\)

d,

= lim_x->2   \(\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x-1\right)}\)

= lim_x->2   \(\dfrac{\left(x+2\right)}{\left(x-1\right)}\)

= 4