Question

Lim(e^(sinx)-e^(sin2x))/2 при х стремящемся к нулю. решить не используя правило лапиталя, не дифференцируя.

Answer 1

$\lim_{x \to 0} \frac{e^{sinx} - e^{sin2x}}{2} = \frac{e^{sin0} - e^{sin0}}{2} = \frac{1-1}{2} = 0$
$II.-\lim_{x \rightarrow 0}\frac{e^{\sin 2x}-e^{\sin x}}{x}=-\lim_{x \rightarrow 0}\frac{e^{2\sin x\cos x}-e^{\sin x}}{x}=$
$-\lim_{x \rightarrow 0}\frac{e^{\sin x}(e^{2\sin x\cos x - \sin x}-1)}{x}=$
$-\lim_{x \rightarrow 0}\frac{e^{\sin x(2\cos x -1)}-1}{\sin x(2\cos x -1)}\cdot \frac{\sin x}{x}\cdot e^{\sin x}(2\cos x-1)=$
$-1\cdot1\cdot1\cdot(2-1)=-1$

Answer 2

Calculation of the Limit

To find the limit of the expression lim(e^(sinx)-e^(sin2x))/2 as x approaches zero without using L'Hôpital's rule or differentiation, we can use the properties of trigonometric functions and the exponential function.

Let's break down the expression step by step:

1. Start with the expression lim(e^(sinx)-e^(sin2x))/2. 2. We can rewrite e^(sin2x) as (e^(sinx))^2. 3. Now, the expression becomes lim(e^(sinx)-(e^(sinx))^2)/2.

To proceed further, we can use the fact that sinx is approximately equal to x for small values of x. This approximation is known as the small-angle approximation.

4. Apply the small-angle approximation to the expression: lim(e^x-(e^x)^2)/2.

Now, let's evaluate the limit using this simplified expression.

5. Substitute x = 0 into the expression: lim(e^0-(e^0)^2)/2. 6. Simplify the expression: lim(1-1)/2. 7. Evaluate the limit: lim(0)/2 = 0.

Therefore, the limit of the expression (e^(sinx)-e^(sin2x))/2 as x approaches zero is 0.

Conclusion

The limit of the expression (e^(sinx)-e^(sin2x))/2 as x approaches zero is 0. This result was obtained by using the small-angle approximation for sinx and evaluating the limit directly without using L'Hôpital's rule or differentiation.

Please note that this solution is based on the given expression and the properties of trigonometric and exponential functions.