Evaluate the Limit

[nycmathguy]

Calculus
Section 2.3

Note: The following question is not covered by Jenn in any of her limits video lessons. However, I am curious about how this one is done.

I am interested in number 66. Clearly, I must rationalize the top or bottom as step 1, but which one? If the limit DNE, explain why.

 

[MathLover1]

66.

lim((sqrt(6-x)-2)/(sqrt(3-x)-1), x->2)=(sqrt(6-2)-2)/(sqrt(3-2)-1)=(sqrt(4)-2)/(sqrt(1)-1)=(2-2)/(1-1)=0/0 which is an undefined form and we can apply L'Hopital rule

hence we get

lim((sqrt(6-x)-2)/(sqrt(3-x)-1), x->2)

=lim(d(sqrt(6-x)-2)/dx)/((sqrt(3-x)-1)/dx), x->2)…

=lim((-1/(2sqrt(6-x)))/(-1/(2sqrt(3-x))), x->2

=lim(sqrt(3-x)/sqrt(6-x), x->2)

=1/2

 

[nycmathguy]

66.

lim((sqrt(6-x)-2)/(sqrt(3-x)-1), x->2)=(sqrt(6-2)-2)/(sqrt(3-2)-1)=(sqrt(4)-2)/(sqrt(1)-1)=(2-2)/(1-1)=0/0 which is an undefined form and we can apply L'Hopital rule

hence we get

lim((sqrt(6-x)-2)/(sqrt(3-x)-1), x->2)

=lim(d(sqrt(6-x)-2)/dx)/((sqrt(3-x)-1)/dx), x->2)…

=lim((-1/(2sqrt(6-x)))/(-1/(2sqrt(3-x))), x->2

=lim(sqrt(3-x)/sqrt(6-x), x->2)

=1/2

Very good but please use wolfram or LaTex for your reply moving forward. Better yet, write your reply on paper, take a picture with your cell phone and upload your answer.

 

[MathLover1]

evaluate for x=2


=

=

=

=

which is an undefined form and we can apply L'Hopital rule

hence we get

=> derivative of

and

so, we have to find lim of

, simplified will be (continues)

 

[MathLover1]

and

 

[nycmathguy]

View attachment 2971 evaluate for x=2


=View attachment 2972

=View attachment 2973

=View attachment 2974

=View attachment 2975 which is an undefined form and we can apply L'Hopital rule

hence we get
View attachment 2971

=> derivative of View attachment 2976 and View attachment 2977

so, we have to find lim ofView attachment 2980 , simplified will be (continues)

Fabulous work! Thank you.

 

[nycmathguy]

View attachment 2981

and View attachment 2982

Amazing simplified work. The limit is 1/2. Thank you.