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Limits: Let Cases 4pt 4pt Cases Cases 4pt 4pt

JEE Maths question with a full step-by-step solution.

Question
Let f(x)={xtan⁡x,−∞<x<02,x=0ln⁡(1+2x)x,0<x<∞f(x)=\begin{cases}\dfrac{x}{\tan x}, & -\infty<x<0\\[4pt] 2, & x=0\\[4pt] \dfrac{\ln(1+2x)}{x}, & 0<x<\infty\end{cases} and g(x)={x+4,−∞<x<1x2−5x+11,1≤x<2x−3,2≤x≤∞g(x)=\begin{cases}x+4, & -\infty<x<1\\[4pt] x^{2}-5x+11, & 1\le x<2\\[4pt] x-3, & 2\le x\le\infty\end{cases} which of the following is(are) correct?
Alim⁡x→0−g(f(x))=5\displaystyle\lim_{x\to 0^{-}}g(f(x))=5correct
Blim⁡x→0−g(f(x))=7\displaystyle\lim_{x\to 0^{-}}g(f(x))=7
Clim⁡x→0+g(f(x))=5\displaystyle\lim_{x\to 0^{+}}g(f(x))=5correct
Dlim⁡x→0+g(f(x))=−1\displaystyle\lim_{x\to 0^{+}}g(f(x))=-1
Solution
Step 1: Compute lim⁡x→0−f(x)\displaystyle\lim_{x\to 0^{-}}f(x). For x<0x<0: f(x)=xtan⁡xf(x)=\dfrac{x}{\tan x}. Let h=−x→0+h=-x\to 0^{+}:
lim⁡h→0+−htan⁡(−h)=lim⁡h→0+−h−tan⁡h=lim⁡h→0+htan⁡h=1\lim_{h\to 0^{+}}\dfrac{-h}{\tan(-h)}=\lim_{h\to 0^{+}}\dfrac{-h}{-\tan h}=\lim_{h\to 0^{+}}\dfrac{h}{\tan h}=1
 So as x→0−x\to 0^{-}, f(x)→1f(x)\to 1, but slightly less than 11 (since htan⁡h<1\dfrac{h}{\tan h}<1 for small h>0h>0). Step 2: Compute lim⁡x→0+f(x)\displaystyle\lim_{x\to 0^{+}}f(x). For x>0x>0: f(x)=ln⁡(1+2x)x=ln⁡(1+2x)2x⋅2→1⋅2=2f(x)=\dfrac{\ln(1+2x)}{x}=\dfrac{\ln(1+2x)}{2x}\cdot 2\to 1\cdot 2=2. So as x→0+x\to 0^{+}, f(x)→2f(x)\to 2, slightly less than 22. Step 3: Compute lim⁡x→0−g(f(x))\displaystyle\lim_{x\to 0^{-}}g(f(x)). As x→0−x\to 0^{-}, f(x)→1−f(x)\to 1^{-} (just below 11). Use the first piece of gg (since input <1<1):
g(f(x))=f(x)+4→1+4=5g(f(x))=f(x)+4\to 1+4=5
 Option (1) is correct. Step 4: Compute lim⁡x→0+g(f(x))\displaystyle\lim_{x\to 0^{+}}g(f(x)). As x→0+x\to 0^{+}, f(x)→2−f(x)\to 2^{-} (just below 22). Use the middle piece of gg (since input is in [1,2)[1,2)):
g(f(x))=f(x)2−5f(x)+11→4−10+11=5g(f(x))=f(x)^{2}-5f(x)+11\to 4-10+11=5
 Option (3) is correct. Correct answers: (1) and (3)
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