Question

limit x tends to 0 e^sinx - 1 / x?

Answer 1

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Answer 2

Given:

e^sinx - 1 / x

To Find:

limit x tends to 0 e^sinx - 1 / x.

Solution:

We need to find :

• $lim_{x->0}$ $\frac{e^{sinx} -1 }{x}$ .

If we directly put x =0 , we can see that the expression will become 0/0 form.

Here we can use the L'Hospital Rule.

• We can find the limit of derivative of numerator and denominator.
• Derivative of numerator:     cos x $e^{sinx}$
• Derivative of Denominator : 1
• Therefore ,
• $lim_{x->0}$ $\frac{e^{sinx} -1 }{x}$ =  $lim_{x->0}$cos x $e^{sinx}$  = 1 .

The limit x tends to 0 e^sinx - 1 / x is 1.