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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Limits

NDA 2015 · Mathematics · Limits

If f(x)= (e^ x-2 -1) (x-1) , then _ x arrow 2 f(x) is equal to

If f(x)=sin⁡(ex−2−1)ln⁡(x−1)f(x)=\frac{\sin \left(e^{x-2}-1\right)}{\ln (x-1)}, then lim⁡x→2f(x)\lim _{x \rightarrow 2} f(x) is equal to

Options

  1. A-2
  2. B-1
  3. C0
  4. D1

Correct answer

D. 1

Step-by-step solution

f(x)=sin⁡(ex−2−1)ln⁡(x−1)lim⁡x→2sin⁡(ex−2−1)ln⁡(x−1)=L\begin{aligned} & f(x)=\frac{\sin \left(e^{x-2}-1\right)}{\ln (x-1)} \\ & \lim _{x \rightarrow 2} \frac{\sin \left(e^{x-2}-1\right)}{\ln (x-1)}=L \end{aligned} It is 00\frac{0}{0} (undefined) condition so using L'hospital's rule ⇒L=lim⁡x→2[{sin⁡(\mathrmex−2−1)}−{ln⁡(x−1)}−]⇒L=lim⁡x→2cos⁡(\mathrmex−2−1)⋅\mathrme(x−2)1/(x−1)⇒L=lim⁡x→2cos⁡(\mathrme2−2−1)\mathrme2−2⋅(2−1)⇒L=cos⁡(0)e0⋅1⇒L=1\begin{aligned} & \Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2}\left[\frac{\left\{\sin \left(\mathrme^{\mathrm{x}-2}-1\right)\right\}^{-}}{\{\ln (\mathrm{x}-1)\}^{-}}\right] \\ & \Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2} \frac{\cos \left(\mathrme^{\mathrm{x}-2}-1\right) \cdot \mathrme^{(\mathrm{x}-2)}}{1 /(\mathrm{x}-1)} \\ & \Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2} \cos \left(\mathrme^{2-2}-1\right) \mathrme^{2-2} \cdot(2-1) \\ & \Rightarrow \mathrm{L}=\cos (0) \mathrm{e}^0 \cdot 1 \\ & \Rightarrow \mathrm{ L}=1 \end{aligned}
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