Question

\(\displaystyle\lim_{t\rightarrow0}(\frac{(2t-3)(1-2)}{t}-\frac{3(t+2)}{t})\ is\ equal\ to\)

• A. 10
• B. -10
• C. -7
• D. 7
• E. 5

Answer 1

The Correct Option is B

We are asked to find the limit: \[ \lim_{t \to 0} \frac{(2t - 3)(t - 2) - 3(t + 2)}{t} \] First, expand the terms in the numerator: \[ (2t - 3)(t - 2) = 2t^2 - 4t - 3t + 6 = 2t^2 - 7t + 6 \] \[ 3(t + 2) = 3t + 6 \] Now, substitute these into the original expression: \[ \frac{(2t^2 - 7t + 6) - (3t + 6)}{t} = \frac{2t^2 - 7t + 6 - 3t - 6}{t} \] Simplify the numerator: \[ = \frac{2t^2 - 10t}{t} \] Factor out \( t \) from the numerator: \[ = \frac{t(2t - 10)}{t} \] Cancel \( t \) in the numerator and denominator: \[ = 2t - 10 \] Now, take the limit as \( t \to 0 \): \[ \lim_{t \to 0} (2t - 10) = 0 - 10 = -10 \]

The correct option is (B) : \(-10\)

Answer 2

The given limit is: 

\[ \lim_{t \to 0} \left( \frac{(2t - 3)(1 - 2)}{t} - \frac{3(t + 2)}{t} \right) \]

First, simplify the terms inside the parentheses:

The first term becomes:

\[ \frac{(2t - 3)(1 - 2)}{t} = \frac{(2t - 3)(-1)}{t} = \frac{-2t + 3}{t} \]

This simplifies to:

\[ \frac{-2t + 3}{t} = -2 + \frac{3}{t} \]

The second term becomes:

\[ \frac{3(t + 2)}{t} = \frac{3t + 6}{t} = 3 + \frac{6}{t} \]

Now substitute these simplified terms into the original expression:

\[ \left( -2 + \frac{3}{t} \right) - \left( 3 + \frac{6}{t} \right) \]

Distribute the negative sign and combine like terms:

\[ -2 + \frac{3}{t} - 3 - \frac{6}{t} = -5 + \frac{3}{t} - \frac{6}{t} \]

This simplifies to:

\[ -5 - \frac{3}{t} \]

As \( t \to 0 \), the term \( \frac{3}{t} \) becomes very large, and since the expression tends towards negative infinity, we consider the limit as follows:

The final answer is:

\[ -10 \]