Question

\(\displaystyle\lim_{t\rightarrow0}\frac{sin(t^2)}{tsin(5t)} \ is\ equal\ to\)

• A. 5
• B. 25
• C. \(\frac{1}{25}\)
• D. \(\frac{1}{5}\)
• E. 0

Answer 1

The Correct Option is D

We are tasked with evaluating the following limit: \[ \lim_{t \to 0} \frac{\sin(t^2)}{t \sin(5t)}. \]
Step 1: Simplify the expression First, notice that as \( t \to 0 \), both \( \sin(t^2) \) and \( \sin(5t) \) approach 0, making the limit of the form \( \frac{0}{0} \), which suggests the use of L'Hopital's Rule. However, before using L'Hopital's Rule, let's rewrite the expression to simplify the computation. We know the following standard limit: \[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1. \]
Step 2: Apply the standard limit We can split the given expression as follows: \[ \frac{\sin(t^2)}{t \sin(5t)} = \frac{\sin(t^2)}{t^2} \cdot \frac{t^2}{t \sin(5t)}. \] Now, evaluate each part separately: 1. \(\lim_{t \to 0} \frac{\sin(t^2)}{t^2} = 1\) (since \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \) and we have \( t^2 \) instead of \( t \)), 2. \(\lim_{t \to 0} \frac{t^2}{t \sin(5t)} = \lim_{t \to 0} \frac{t}{\sin(5t)} \cdot \lim_{t \to 0} \frac{1}{5} = \frac{1}{5}\).
Step 3: Combine the results So the overall limit becomes: \[ 1 \cdot \frac{1}{5} = \frac{1}{5}. \]

The correct option is (D) : \(\frac{1}{5}\)

Answer 2

We want to evaluate the limit: \[\displaystyle\lim_{t\rightarrow0}\frac{\sin(t^2)}{t\sin(5t)}\]

1. We can rewrite the expression using the fact that \(\displaystyle\lim_{x\rightarrow0} \frac{\sin(x)}{x} = 1\).
2. Rewrite the limit as: \[\displaystyle\lim_{t\rightarrow0}\frac{\sin(t^2)}{t^2} \cdot \frac{5t}{\sin(5t)} \cdot \frac{t^2}{t \cdot 5t} \]
3. Simplify the expression: \[\displaystyle\lim_{t\rightarrow0}\frac{\sin(t^2)}{t^2} \cdot \frac{5t}{\sin(5t)} \cdot \frac{1}{5} \]
4. Now, we can apply the limit: \[\displaystyle\lim_{t\rightarrow0}\frac{\sin(t^2)}{t^2} = 1\] \[\displaystyle\lim_{t\rightarrow0}\frac{5t}{\sin(5t)} = 1\]
5. Thus, the overall limit is: \[1 \cdot 1 \cdot \frac{1}{5} = \frac{1}{5}\]
6. Therefore, the limit is \(\frac{1}{5}\).