Solve f(x) = 2x + 1 for x < 0, x^2 + 1 for x >= 0, limit as x -> 0 Step by Step

Step by step

The question

Question: find the limit as x → 0 of this f(x): 2x + 1 if x < 0, and x² + 1 if x ≥ 0.

f(x) = 2x + 1 if x < 0

= x² + 1 if x ≥ 0

1 From the left

From the left means x is a bit less than 0.

There, f(x) is 2x + 1.

lim x → 0⁻ f(x) = lim x → 0⁻ (2x + 1)

So we use the first piece.

x < 0: piece 1

A polynomial piece has no gaps, so we just put 0 in.

We get 1.

= 2(0) + 1 = 0 + 1 = 1

2 From the right

From the right means x is a bit more than 0.

There, f(x) is x² + 1.

lim x → 0⁺ f(x) = lim x → 0⁺ (x² + 1)

So we use the second piece.

x > 0: piece 2

A polynomial piece has no gaps, so we just put 0 in.

We get 1.

= 0² + 1 = 0 + 1 = 1

3 Compare the sides

From the left we get 1.

From the right we get 1.

left: 1 right: 1

Both sides give the same number.

So the limit is 1.

lim x → 0 f(x) = 1

Careful!

One side is not enough.

A limit needs both sides to give the same number.

Careful: both sides must give the same number

4 Check

Let's check on the graph.

The first piece comes in from the left.

It reaches height 1.

The second piece starts at height 1.

They meet: no jump.

At x = 0 both pieces arrive at 1.

Right!

Let's look back.

First: from the left.

Then: from the right.

Then: compare the sides.

Last: check.

Answer