Question

If $[t]$ represents the greatest integer $\leq t$ then the value of $\lim_{x\to 3} \frac{11-[2-x]}{[x+10]}$ is

• A. 1
• B. 8
• C. 5
• D. does not exist

Hint:

When evaluating limits involving the greatest integer function $[x]$, always check the left-hand and right-hand limits separately. The value of $[f(x)]$ can change depending on which side you approach the limit point from.

Answer 1

The Correct Option is A

To evaluate the limit, we need to find the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL) as $x$ approaches 3.
Step 1: Calculate the Left-Hand Limit (LHL).
Let $x \to 3^-$. This means $x$ is slightly less than 3, for example, $x=2.9$.
Numerator: $2-x \to 2-2.9 = -0.9$. So, $[2-x] = -1$. The numerator becomes $11 - (-1) = 12$.
Denominator: $x+10 \to 2.9+10 = 12.9$. So, $[x+10] = 12$.
LHL = $\lim_{x\to 3^-} \frac{11-[2-x]}{[x+10]} = \frac{12}{12} = 1$.
Step 2: Calculate the Right-Hand Limit (RHL).
Let $x \to 3^+$. This means $x$ is slightly greater than 3, for example, $x=3.1$.
Numerator: $2-x \to 2-3.1 = -1.1$. So, $[2-x] = -2$. The numerator becomes $11 - (-2) = 13$.
Denominator: $x+10 \to 3.1+10 = 13.1$. So, $[x+10] = 13$.
RHL = $\lim_{x\to 3^+} \frac{11-[2-x]}{[x+10]} = \frac{13}{13} = 1$.
Step 3: Compare LHL and RHL.
Since LHL = RHL = 1, the limit exists and its value is 1.