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Solve f(x) = 3x for x < 1, x + 5 for x >= 1, limit as x -> 1

Step by step, with the reason for every step · One-sided limits · Calculus

Step by step

The question

Question: find the limit as x → 1 of this f(x): 3x if x < 1, and x + 5 if x ≥ 1.

f(x) = 3x if x < 1
= x + 5 if x ≥ 1

1 From the left

From the left means x is a bit less than 1.

There, f(x) is 3x.

lim x → 1⁻ f(x) = lim x → 1⁻ 3x

So we use the first piece.

x < 1: piece 1

A polynomial piece has no gaps, so we just put 1 in.

We get 3.

= 3(1) = 3

2 From the right

From the right means x is a bit more than 1.

There, f(x) is x + 5.

lim x → 1⁺ f(x) = lim x → 1⁺ (x + 5)

So we use the second piece.

x > 1: piece 2

A polynomial piece has no gaps, so we just put 1 in.

We get 6.

= 1 + 5 = 6

3 Compare the sides

From the left we get 3.

From the right we get 6.

left: 3 right: 6

The two sides do not agree.

So the limit does not exist.

lim x → 1 f(x) = does not exist

Careful!

One side is not enough.

3 from the left and 6 from the right do not agree.

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 3.

The second piece starts at height 6.

There is a jump.

At x = 1 the graph jumps from 3 to 6.

Right!

Let's look back.

First: from the left.

Then: from the right.

Then: compare the sides.

Last: check.

Answer
Careful: both sides must give the same number

Your turn

Try this one yourself, then check:

Find the limit as f(x) = 6x if x < 3; 3x + 12 if x ≥ 3; x → 3
Show the answer
limit does not exist

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