Question

Lim x→7 кв. корень х - 3-2/х-7

Answer 1

lim(x→7) (√(x-3)-2)/(x-7)

Это неопределённость 0/0:   ⇒

Берём одновременно производную от числителя и знаменателя:

Числитель: (√(x-3)-2)`=1/(2*√(x-3)).

Знаменатель: (x-7)`=1.     ⇒

lim(x→7) (1/(2*√(x-3))=1/(2*√(7-3))=1/(2*√4)=1/(2*2)=1/4.

Answer 2

To solve this limit, we can first simplify the expression by rationalizing the denominator of the second term.

So, we have:

lim x→7 [√(x) - 3 - 2/(x - 7)]

Multiplying the second term by (x + 7)/(x + 7), we get:

lim x→7 [√(x) - 3 - 2(x + 7)/((x - 7)(x + 7))]

Simplifying the second term further, we get:

lim x→7 [√(x) - 3 - (2x + 14)/(x^2 - 49)]

Now, we can combine the first two terms by finding a common denominator:

lim x→7 [(√(x)(x^2 - 49) - 3(x^2 - 49) - (2x + 14))/(x^2 - 49)]

Simplifying the numerator, we get:

lim x→7 [(x√(x)^2 - 3x^2 + 147 - 2x - 14)/(x^2 - 49)]

lim x→7 [(x - 7)(√(x)^2 + 7√(x) + 21)/(x^2 - 49)]

Now, we can cancel out the common factor of (x - 7) in the numerator and denominator, and simplify further:

lim x→7 [(√(x)^2 + 7√(x) + 21)/(x + 7)(x - 7)]

Substituting x = 7, we get:

(√(7)^2 + 7√(7) + 21)/(7 + 7)(7 - 7)

= (7 + 7√(7) + 21)/0

This limit is undefined since we cannot divide by zero. Therefore, the limit does not exist.