Question

1) lim(х^2-3x+5) x-->1 2) lim(x^2+6x-8) x-->-1

Answer 1

1) 1.1- 3.1 + 5 =1-3+5=3
2)(-1).(-1)+ 6.(-1) -8 = 1-6-8 = -13

Answer 2

В обоих случаях надо просто подставить вместо х его значение.
в первом примере предел равен 3, во втором -13

Answer 3

To find the limits of the given functions as x approaches certain values, you can simply plug in the value of x into the functions and evaluate them. Let's calculate the limits:

1. $\lim_{{x \to 1}} (x^2 - 3x + 5)$ Replace x with 1: $(1^2 - 3(1) + 5) = (1 - 3 + 5) = 3$

So, $\lim_{{x \to 1}} (x^2 - 3x + 5) = 3$.

2. $\lim_{{x \to -1}} (x^2 + 6x - 8)$ Replace x with -1: $((-1)^2 + 6(-1) - 8) = (1 - 6 - 8) = -13$

So, $\lim_{{x \to -1}} (x^2 + 6x - 8) = -13$.