Question

\(\lim_{x\to0} e^{x}-\dfrac{1}{3}×(1-e^{2x})\)\(=\)?

• A. \(\dfrac{-1}{6}\)
• B. \(\dfrac{1}{6}\)
• C. \(1\)
• D. \(0\)
• E. \(\dfrac{-1}{3}\)

Answer 1

The Correct Option is C

\(\lim_{x\to0} e^{x}-\dfrac{1}{3}×(1-e^{2x})\)

Now substituting  \(X=0\) and on simplification we get,

\(= e^0-\dfrac{1}{3}(1-e^{2.0})\)

 \(=1-0\)

\(=1\)  (Ans)

 

Answer 2

\(\lim_{x \to 0} e^{x} - \frac{1}{3} \times (1 - e^{2x})\)

Let's simplify the expression inside the parentheses:

\(1 - e^{2x} = 1 - (e^x)^2\)

Now, as x approaches 0:

\(\lim_{x \to 0} e^{x} - \frac{1}{3} \times (1 - e^{2x}) = \lim_{x \to 0} e^{x} - \frac{1}{3} \times (1 - (e^x)^2)\)

Now, let's factor out \(e^x\) from the expression:

\(= \lim_{x \to 0} e^{x} - \frac{1}{3} \times (1 - e^{x}) \times (1 + e^{x})\)

Now, as x approaches 0:

\(= e^0 - \frac{1}{3} \times (1 - e^0) \times (1 + e^0)\)

\(= 1 - \frac{1}{3} \times (1 - 1) \times (1 + 1)\)

\(= 1 - \frac{1}{3} \times (0) \times (2)\)

\(= 1 - 0 \times 2\)

\(= 1\)

So, the correct option is (C): 1