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Worked Examples · Example 13

Q.Discuss the continuity of the function ff given by f(x)={x,if x≥0x2,if x<0f(x) = \begin{cases} x, & \text{if } x \geq 0 \\ x^2, & \text{if } x < 0 \end{cases}.

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At x=0x=0 the two one-sided limits and f(0)f(0) all equal 00, so ff is continuous there; each piece is a polynomial, so ff is continuous on all of R\mathbb{R}.

Where to look

Continuity at x=ax=a needs f(a)f(a) defined, lim⁡x→af(x)\lim_{x\to a}f(x) to exist, and the two to be equal. The pieces xx and x2x^2 are polynomials, continuous on their own, so only the join at x=0x=0 needs checking.

Value at the point

Since 0≥00\ge 0, the top rule applies: f(0)=0f(0)=0.

One-sided limits

From the left (x<0x<0, f(x)=x2f(x)=x^2):

lim⁡x→0−f(x)=02=0.\lim_{x\to 0^-} f(x) = 0^2 = 0.

From the right (x>0x>0, f(x)=xf(x)=x):

lim⁡x→0+f(x)=0.\lim_{x\to 0^+} f(x) = 0.

Compare

Both one-sided limits are 00, so lim⁡x→0f(x)=0\lim_{x\to 0}f(x)=0, and this equals f(0)=0f(0)=0. All three continuity conditions hold at x=0x=0.

Watch out

Two different formulas (xx vs x2x^2) do not force a jump. Compute the limits — here both sides meet at 00, so there is no break. …

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