Question

Please do it kindly

Answer 1

Solution

Find :-

• lim x ->0 [ e^x - e^(x cos x)]/(x + sin x)

Explanation

We know, If lim of numerator & denominator is 0/0 .

In this condition we can use " L'Hospital Rule " .

L'Hospital Rule " .

If lim x ->0 [ p/q] So , by L'Hospital rule it be = lim x ->[ (dp/dx)/(dq/dx)]

So, Applying here,

Now, Calculate Differential of e^x - e^(x cos x)]/(x + sin x)

==> lim x ->0 [ d{e^x - e^(x cos x)}/dx]/d(x + sin x)/dx

Some Important differential

★ d e^x/dx = e^x

★ d( sin x)/dx = cos x

★ d( cos x)/dx = - sin x

★ d(pq)/dx = q *dp/dx + p * dq/dx

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So Now,

==> lim x->0 [ e^x - e^(x cos x) (x * - sin x + cos x )]/(1+cos x)

Used Formula

★ e^0 = 1

★ Sin 0 = 0

★ cos 0 = 1

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So,

==> [e^0 - e^(0*cos 0)(0* - sin 0 + cos 0)]/(1+ cos 0)

==> [1 - e^(0)(0+1)]/(1+1)

==> (1- 1 * 1 * 2)/(1 + 1)

==> ( 1 - 2 )/2

==> -1/2

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Hence

• Value of lim x ->0 [ e^x - e^(x cos x)]/(x + sin x) = -1/2

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