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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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