Exercise 5.1 · Q18: $\lambda(x^2-2x)$ and $4x+1$⁠: continuity at $0$ and $1$

Basics you need

Part (i): continuity at $x=0$

1. $0$ is in the domain; find $f(0)$

   Every real $x$ satisfies exactly one of "$x\le0$" and "$x\gt0$", so $f$ is defined on all of $\R$⁠. Since $0\le0$⁠, the first rule gives

   $$f(0)=\lambda\left(0^2-2\cdot0\right)=\lambda\cdot0=0.$$

   This is $0$ for every value of $\lambda$⁠.

   The value at the junction comes from the line whose condition contains the junction, here "$x\le0$". The constant $\lambda$ multiplies a bracket that vanishes at $0$⁠, so it cannot affect $f(0)$⁠.
   Reading a piecewise rule: the condition decides the formula. Basics: piecewise functions

2. The left-hand limit at $0$

   Put $x=-h$ with $h\gt0$⁠, $h\to0$⁠. Then $x\lt0$⁠, so the first rule applies. Since $(-h)^2=h^2$ and $-2(-h)=2h$⁠,

   $$f(-h)=\lambda\left(h^2+2h\right).$$

   As $h\to0$⁠, $h^2+2h\to0$⁠, so

   $$\lim_{x\to0^-}f(x)=\lim_{h\to0}\lambda\left(h^2+2h\right)=\lambda\cdot0=0.$$
   Take care with the signs: squaring removes the minus, but $-2x$ at $x=-h$ becomes $+2h$⁠. Either way both terms shrink to $0$⁠, and multiplying by the fixed number $\lambda$ keeps the limit at $0$⁠.
   For a fixed constant $\lambda$⁠, $\lim\lambda g(h)=\lambda\lim g(h)$⁠. Basics: left-hand and right-hand limits

3. The right-hand limit at $0$

   Put $x=h$ with $h\gt0$⁠, $h\to0$⁠. Now $x\gt0$⁠, so the second rule applies:

   $$\begin{aligned} f(h)&=4h+1,\\ \lim_{x\to0^+}f(x)&=\lim_{h\to0}\,(4h+1)=1. \end{aligned}$$
   Points just to the right of $0$ are governed by $4x+1$⁠, which heads for height $1$⁠. The value $f(0)=0$ is irrelevant to this limit: a limit never looks at the point itself.
   $4h\to0$ as $h\to0$⁠. Basics: what a limit is

4. Compare: no $\lambda$ can work

   From the three steps above:

   $$\begin{aligned} \lim_{x\to0^-}f(x)&=0,\\ f(0)&=0,\\ \lim_{x\to0^+}f(x)&=1. \end{aligned}$$

   Continuity at $0$ would need $0=1$⁠, which is false. None of the three numbers involves $\lambda$⁠, so the failure happens for every value of $\lambda$. Hence there is no value of $\lambda$ for which $f$ is continuous at $x=0$.

   The question "for what value of $\lambda$" invites us to solve an equation in $\lambda$⁠. But the equation we get, $0=1$⁠, has no $\lambda$ in it and is never true, so the honest answer is "none". (The left-hand limit does equal $f(0)$⁠, so $f$ is continuous from the left at $0$⁠; it is the right side that jumps.)
   A jump (left-hand limit $\ne$ right-hand limit) is a discontinuity, whatever the value at the point. Lesson: continuity at a point

Part (ii): continuity at $x=1$

1. Which rule governs the points near $1$⁠?

   $1$ is in the domain and $1\gt0$⁠, so

   $$f(1)=4(1)+1=5.$$

   Moreover every $x$ with $0\lt x\lt2$ satisfies $x\gt0$⁠, so on the whole open interval $(0,2)$ around $1$ we have $f(x)=4x+1$⁠. The rule containing $\lambda$ is used only for $x\le0$⁠, at distance at least $1$ from the point $1$⁠.

   A limit at $1$ only uses values of $f$ at points as close to $1$ as we like. Once we stay within distance $1$ of the point, the $\lambda$-rule never appears, so $\lambda$ cannot influence continuity at $1$⁠.
   Continuity is a local property: it depends only on the function near the point. Basics: what a limit is

2. The two one-sided limits at $1$

   Take $0\lt h\lt1$⁠, so that $1-h$ and $1+h$ are both in $(0,2)$⁠:

   $$\begin{aligned} \lim_{x\to1^-}f(x)&=\lim_{h\to0}\,\left[4(1-h)+1\right]\\ &=\lim_{h\to0}\,(5-4h)=5,\\ \lim_{x\to1^+}f(x)&=\lim_{h\to0}\,\left[4(1+h)+1\right]\\ &=\lim_{h\to0}\,(5+4h)=5. \end{aligned}$$
   The restriction $h\lt1$ guarantees $1-h\gt0$⁠, so the left-hand points really are governed by $4x+1$⁠. Since $h\to0$⁠, only small $h$ matter anyway.
   One-sided limits computed with $x=1\mp h$⁠. Basics: left-hand and right-hand limits

3. Compare

   Left-hand limit $=5$⁠, right-hand limit $=5$⁠, $f(1)=5$⁠. All three agree, so $f$ is continuous at $x=1$ for every value of $\lambda$.

   This is Definition 1 applied directly. Equivalently: near $1$⁠, $f$ coincides with the polynomial $4x+1$⁠, and polynomials are continuous (Example 14).
   Polynomials are continuous at every real number. Lesson: continuity at a point

Every fact used, and where it comes from