To determine

To find: Evaluate limx→πesin(x)−1x−π .

Answer

−1

Explanation

Given: limx→πesin(x)−1x−π

Formula used:

L’Hospital’s rule:

Suppose limx→af(x)g(x)=00 or limx→af(x)g(x)=±∞±∞ then limx→af(x)g(x)=limx→af′(x)g′(x); where a is any real number.

limx→πesin(x)−1x−π=limy→0esin(y+π)−1y=limy→0esin(y)−1y=limy→0e−sin(y)−1−siny(−sin(y)y)=−limu→0eu−1ulimy→0sin(y)y

As per definition, first limit is equal to 1,

Therefore,

limx→πesin(x)−1x−π=−limy→0sin(y)y=−1

Conclusion: Hence, the value of limit is −1.