Q.A mobile-recharge plan charges $f(x) = 2x$ (rupees) for $x$ minutes of talk-time when $x \le 10$, and $f(x) = 3x - 10$ (rupees) when $x > 10$. Find the left-hand limit and the right-hand limit of $f(x)$ as $x \to 10$, and state whether $\lim_{x\to10} f(x)$ exists.

For $x$ slightly less than $10$, the plan uses $f(x)=2x$; for $x$ slightly more than $10$, it uses $f(x)=3x-10$ — we evaluate each piece separately at $x=10$.

LHL: $\lim_{x\to10^-} 2x = 2(10)=20$. RHL: $\lim_{x\to10^+} (3x-10) = 3(10)-10=20$. Since LHL $=$ RHL $=20$, the limit exists and equals $20$.

Step 1 — Left-hand limit. As $x\to10^-$ (values just below $10$, e.g. $9.9, 9.99,\ldots$), the plan is governed by $f(x)=2x$ (since these values satisfy $x\le10$). So

Step 2 — Right-hand limit. As $x\to10^+$ (values just above $10$, e.g. $10.1, 10.01,\ldots$), the plan is governed by $f(x)=3x-10$ (since these values satisfy $x>10$). So

Step 3 — Compare. LHL $=20$ and RHL $=20$ — they agree, so by the existence rule (§2), the two-sided limit exists:

Cross-check: notice this also means the recharge charge is continuous at the $10$-minute breakpoint (no sudden jump in price right at $10$ minutes) — a sensible design for a real tariff plan, which is exactly why the two pieces were chosen to meet at $x=10$.

Direct check by continuity: since both pieces are linear and agree in value at $x=10$ (piece 1 gives $2(10)=20$, piece 2 gives $3(10)-10=20$), the function is continuous there, which independently confirms the limit exists and equals $f(10)=20$.

A common mistake is plugging $x=10$ into the WRONG piece for one of the one-sided limits — e.g. using $3x-10$ for the LHL because $10$ itself falls in the $x\le10$ piece, forgetting that the LHL is about values strictly LESS than $10$ approaching it, which is exactly where the $x\le10$ piece (i.e. $2x$) applies.