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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Limits

NTA Abhyas 2020 · Mathematics · Limits

The value of _ x → 1 ^ + ∫ _ 1 ^ x |t - 1|dt ( x - 1 ) is equal to

The value oflim⁡x→1+∫1x∣t−1∣dtsin⁡(x−1)\lim_{x \to {1}^{+}}\dfrac{{∫}_{1}^{x}\left|t - 1\right|dt}{\sin ( x - 1 )} is equal to

Correct answer

0

Step-by-step solution

Given:lim⁡x→1+∫1x(t−1)dtsin⁡(x−1)\lim_{x \to {1}^{+}} \dfrac{{∫}_{1}^{x} ( t - 1 ) dt}{\sin ( x - 1 )}

Using L’Hospital’s rule and leibuitzrule,

l=lim⁡x→1+ddx(∫1x(t−1)dt)ddx(sin⁡(x−1))l = \lim_{x \to {1}^{+}} \dfrac{\dfrac{d}{dx} \left({∫}_{1}^{x} ( t - 1 ) dt\right)}{\dfrac{d}{dx} \left(\sin \left(x - 1\right)\right)}

lim⁡x→1+(x−1)cos⁡(x−1)=01=0\lim_{x \to {1}^{+}} \dfrac{\left(x - 1\right)}{\cos \left(x - 1\right)} = \dfrac{0}{1} = 0
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