Respuesta :

ANCKPOP

Answer:

[tex]\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\  \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right[/tex]

Step-by-step explanation:

a) [tex]\lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}[/tex]

b) [tex]\lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}[/tex]

c) [tex]\lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0[/tex]

d) [tex]\lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}[/tex]

e) [tex]\lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2[/tex]

f) [tex]\lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}[/tex]

ashishdwivedilVT

We find the following limit values using the L Hospital rule.

1)-3/4

2)-11/2

3)28/25

4)-2

5)11/2

6)0

Apply L hospital rule in each expression since it is getting 0/0 form after putting the value of the x, except in the last problem(problem6)

What are the L hospital rules?

L’Hospital’s Rule says that if we get an indeterminate form i.e. 0/0 or ∞/∞ after putting the value of the independent variable in the function, we should differentiate both the numerator as well as the denominator then again apply the limits.

now, we differentiate the numerator as well as the denominator to find the limit.

1) [tex]\displaystyle \lim_{ x\to 3}\frac{x^2-10x+21}{x^2+4x-21}[/tex][tex]=\displaystyle \lim_{ x\to3}\frac{2x-10}{2x+4}[/tex][tex]=-\frac{2}{5}[/tex]

2)[tex]\displaystyle \lim_{ x\to -3/2}\frac{2x^2-5x-12}{2x+3}[/tex] [tex]=\displaystyle \lim_{ x\to-3/2}\frac{4x-5}{2}[/tex] [tex]= -\frac{11}{2}[/tex]

3)[tex]\displaystyle \lim_{ x\to 11}\frac{x^2+6x-187}{x^2+3x-154}[/tex][tex]=\displaystyle \lim_{ x\to11}\frac{2x+6}{2x+3}= \frac{28}{25}[/tex]

4)[tex]\displaystyle \lim_{ x\to 3}\frac{x^2-8x+15}{x-3}[/tex][tex]=\displaystyle \lim_{ x\to3}\frac{2x-8}{1}= \frac{-2}{1}= -2[/tex]

5)[tex]\displaystyle \lim_{ x\to5/2}\frac{2x^2+x-15}{2x-5}[/tex][tex]=\displaystyle \lim_{ x\to 5/2}\frac{4x+1}{2}= \frac{11}{2}[/tex]

for prob6 put x=2 in the expression

6) [tex]\displaystyle \lim_{ x\to2}\frac{x-2}{x^2-2}[/tex][tex]= \frac{2-2}{x^2-2}[/tex][tex]= 0[/tex]

Therefore, we find the following limit values using the L Hospital rule.

1)-3/4

2)-11/2

3)28/25

4)-2

5)11/2

6)0

to get more about the limit refer to the link,

https://brainly.com/question/105479

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