Continuity flash cards

1. 1.Define continuity of $f(x)$ at $x=a$ using the epsilon-delta / limit definition.

   $f$ is continuous at $x=a$ iff $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$, i.e. LHL = RHL = value of function, and all three must exist (finite).

   Hint: Three things must match

2. 2.If $f$ and $g$ are continuous at $x=a$, list which combinations are guaranteed continuous at $a$ and note the one exception.

   $f\pm g$, $f\cdot g$, $cf$, $|f|$, and $f/g$ are continuous at $a$ provided $g(a)\neq0$ for the quotient. Composition $f(g(x))$ is continuous at $a$ if $g$ is continuous at $a$ and $f$ is continuous at $g(a)$.

   Hint: Division needs a nonzero check

3. 3.For a function defined by cases (piecewise) at a junction point $x=a$, what is the standard 3-step test for continuity, and how do you find an unknown constant making it continuous?

   Compute (1) $\lim_{x\to a^-}f(x)$, (2) $\lim_{x\to a^+}f(x)$, (3) $f(a)$. Set all three equal and solve the resulting equation for the unknown constant (often using standard limits like $\lim_{x\to0}\frac{\sin x}{x}=1$).

   Hint: LHL = RHL = f(a)

4. 4.State the Intermediate Value Theorem (IVT) and its most common JEE application.

   If $f$ is continuous on $[a,b]$ and $k$ lies between $f(a)$ and $f(b)$, then there exists $c\in(a,b)$ with $f(c)=k$. Application: if $f(a)\cdot f(b)<0$, then $f$ has at least one root in $(a,b)$.

   Hint: Used to prove existence of roots

5. 5.Give an example distinguishing a removable discontinuity from a non-removable (jump) discontinuity, and how each is characterized in terms of limits.

   Removable: $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)\neq f(a)$ (or undefined at $a$) — can be fixed by redefining $f(a)$. Non-removable (jump): $\lim_{x\to a^-}f(x)\neq\lim_{x\to a^+}f(x)$ — cannot be fixed by redefinition; e.g. $f(x)=\frac{\sin x}{x}$ at $x=0$ (removable) vs. $f(x)=[x]$ (greatest integer, jump) at integer points.

   Hint: Can redefining one point fix it?

6. 6.A function is defined as $f(x)=\dfrac{1-\cos 4x}{x^2}$ for $x<0$, $f(0)=a$, and $f(x)=\dfrac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$ for $x>0$. Find $a$ for continuity at $x=0$.

   As $x\to0^-$: $1-\cos4x=2\sin^2 2x$, so $\dfrac{2\sin^2 2x}{x^2}=8\left(\dfrac{\sin 2x}{2x}\right)^2\to 8$. As $x\to0^+$: rationalising, $\dfrac{\sqrt{x}}{\sqrt{16+\sqrt x}-4}=\sqrt{16+\sqrt x}+4\to 8$. Both one-sided limits equal $8$, so $a=8$.

   Hint: Use $1-\cos4x=2\sin^2 2x$ on the left; rationalise the surd on the right.

7. 7.$f(x)=\dfrac{\sin(a+1)x+\sin x}{x}$ for $x<0$, $f(0)=c$, and $f(x)=\dfrac{\sqrt{x+bx^2}-\sqrt{x}}{bx^{3/2}}$ for $x>0$ ($b\neq0$). Find $a$ and $c$ so that $f$ is continuous at $0$.

   Left limit: $\dfrac{\sin(a+1)x}{x}+\dfrac{\sin x}{x}\to(a+1)+1=a+2$. Right limit: $\dfrac{\sqrt{x}(\sqrt{1+bx}-1)}{bx^{3/2}}=\dfrac{\sqrt{1+bx}-1}{bx}\to\dfrac12$ (independent of $b$). Equating, $a+2=\tfrac12\Rightarrow$ $a=-\dfrac32,\ c=\dfrac12$ (any $b\neq0$ works).

   Hint: Expand both sides near $0$; notice the right-hand limit doesn't depend on $b$.

8. 8.$f(x)=\dfrac{1-\sin^3x}{3\cos^2x}$ for $x<\tfrac{\pi}{2}$, $f\left(\tfrac{\pi}{2}\right)=k$, and $f(x)=\dfrac{b(1-\sin x)}{(\pi-2x)^2}$ for $x>\tfrac{\pi}{2}$. Find $k,b$ for continuity at $\tfrac{\pi}{2}$.

   Put $x=\tfrac{\pi}{2}-h$: $1-\sin^3x=1-\cos^3h\approx\tfrac32h^2$ and $3\cos^2x=3\sin^2h\approx3h^2$, giving left limit $\tfrac12$, so $k=\tfrac12$. Put $x=\tfrac{\pi}{2}+h$: $1-\sin x=1-\cos h\approx\tfrac{h^2}{2}$ and $(\pi-2x)^2=4h^2$, so right limit $=\dfrac{b}{8}$. Equating to $\tfrac12$ gives $k=\dfrac12,\ b=4$.

   Hint: Substitute $x=\pi/2\mp h$ and expand $\sin,\cos$ to second order in $h$.

9. 9.For what value of $k$ is $f(x)=\dfrac{\tan x-\sin x}{x^3}$ ($x\neq0$), $f(0)=k$ continuous at $0$?

   $\tan x-\sin x=\sin x\left(\dfrac{1}{\cos x}-1\right)=\dfrac{\sin x(1-\cos x)}{\cos x}$. Near $0$ this behaves like $\dfrac{x\cdot\frac{x^2}{2}}{1}=\dfrac{x^3}{2}$, so the limit is $\tfrac12$. Hence $k=\dfrac12$.

   Hint: Write $\tan x-\sin x$ as a single fraction over $\cos x$ first.

10. 10.Find $k$ such that $f(x)=\dfrac{1-\cos(1-\cos x)}{x^4}$ for $x\neq0$, $f(0)=k$ is continuous.

    Let $u=1-\cos x\approx\dfrac{x^2}{2}$ for small $x$. Then $1-\cos u\approx\dfrac{u^2}{2}=\dfrac{1}{2}\left(\dfrac{x^2}{2}\right)^2=\dfrac{x^4}{8}$. Dividing by $x^4$ gives the limit $\dfrac18$, so $k=\dfrac18$.

    Hint: Set $u=1-\cos x$ and apply the standard small-angle expansion twice.

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