Find the limit
(calculus)

0

Step-by-step explanation:

Step 1: Define

Step 2: Substitute in infinity

We know that the second exponent is always going to be bigger than the 1st exponent, so we would get a negative exponent.

We can simplify a negative exponent into a fraction.

We know that if we keep plugging in larger numbers for x, the denominator is going to be infinitely big. And bigger the number we are dividing by, the smaller the entire fraction is going to be.

Therefore, we would be approaching 0.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)