Q.Find $\lim_{x\to 0} f(x)$ and $\lim_{x\to 1} f(x)$, where $f(x) = \begin{cases} 2x + 3, & x \le 0 \\ 3(x + 1), & x > 0 \end{cases}$

Concept: Piecewise Limit — check left-hand and right-hand limits separately.

For $\lim_{x\to 0} f(x)$:

• Left-hand limit ($x \to 0^-$): use $2x+3$, so $2(0)+3 = 3$.
• Right-hand limit ($x \to 0^+$): use $3(x+1)$, so $3(0+1) = 3$.
• Both sides equal 3, so the limit exists.

For $\lim_{x\to 1} f(x)$: …

We are given:

The key idea: for a limit to exist at a point, the function must approach the same value from both sides. At $x=0$, the definition changes — so we must check left and right separately. At $x=1$, the function is defined by a single rule (the second piece) in a neighbourhood of $1$, so the limit is just the value of that polynomial.

1. Limit as $x \to 0$

Because the function has different rules on either side of $0$, we compute the left-hand limit and the right-hand limit.

Left-hand limit ($x \to 0^-$):

For $x \le 0$, $f(x) = 2x + 3$. As $x$ approaches $0$ from the left, we substitute directly (since polynomials are continuous):

Right-hand limit ($x \to 0^+$):

For $x > 0$, $f(x) = 3(x + 1)$. As $x$ approaches $0$ from the right:

Both one-sided limits equal $3$. Therefore, the two-sided limit exists and is $3$.

Thus:

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