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Limits Continuity and Differentiability question

2023 · 30 Jan · Shift 2 · Q28
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  5. /2023 · 30 Jan · Shift 2 · Q28

Limits Continuity and Differentiability question

2023 · 30 Jan · Shift 2 · Q28

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f,gf, gf,g and hhh be the real valued functions defined on R\mathbb{R}R as f(x)={x∣x∣,xeq01,x=0g(x)={sin⁡(x+1)(x+1),xeq−11,x=−1f(x)=\left\{\begin{array}{cc}\frac{x}{|x|}, & x eq 0 \\ 1, & x=0\end{array}\right.g(x)=\left\{\begin{array}{cc}\frac{\sin (x+1)}{(x+1)}, & x eq-1 \\ 1, & x=-1\end{array}\right.f(x)={∣x∣x​,1,​xeq0x=0​g(x)={(x+1)sin(x+1)​,1,​xeq−1x=−1​ and h(x)=2[x]−f(x)h(x)=2[x]-f(x)h(x)=2[x]−f(x), where [x][x][x] is the greatest integer ≤x\leq x≤x. Then the value of lim⁡x→1g(h(x−1))\lim\limits_{x \rightarrow 1} g(h(x-1))x→1lim​g(h(x−1)) is :
  1. A
    1
  2. B
    −1-1−1
  3. C
    sin⁡(1)\sin (1)sin(1)
  4. D
    0
View written solutionFree

Correct answer: A

  1. Understand the functions

    We are given:

    f(x)={x∣x∣,x≠01,x=0f(x)= \begin{cases} \dfrac{x}{|x|}, & x\ne 0 \\ 1, & x=0 \end{cases}f(x)=⎩⎨⎧​∣x∣x​,1,​x=0x=0​

    So:

    • if x>0x>0x>0, then f(x)=1f(x)=1f(x)=1
    • if x<0x<0x<0, then f(x)=−1f(x)=-1f(x)=−1
    • and f(0)=1f(0)=1f(0)=1

    Also,

    g(x)={sin⁡(x+1)x+1,x≠−11,x=−1g(x)= \begin{cases} \dfrac{\sin(x+1)}{x+1}, & x\ne -1 \\ 1, & x=-1 \end{cases}g(x)=⎩⎨⎧​x+1sin(x+1)​,1,​x=−1x=−1​

    and

    h(x)=2[x]−f(x)h(x)=2[x]-f(x)h(x)=2[x]−f(x)

    where [x][x][x] denotes the greatest integer function.

  2. We need to find

    lim⁡x→1g(h(x−1))\lim_{x\to 1} g(h(x-1))x→1lim​g(h(x−1))

    Let

    t=x−1t=x-1t=x−1

    Then as x→1x\to 1x→1, we have t→0t\to 0t→0.

    So the limit becomes

    lim⁡t→0g(h(t))\lim_{t\to 0} g(h(t))t→0lim​g(h(t))

  3. Evaluate h(t)h(t)h(t) near t=0t=0t=0

    Since

    h(t)=2[t]−f(t)h(t)=2[t]-f(t)h(t)=2[t]−f(t)

    we consider left and right of 000.

    Case 1: t→0+t\to 0^+t→0+

    For 0<t<10<t<10<t<1,

    [t]=0,f(t)=1[t]=0, \qquad f(t)=1[t]=0,f(t)=1

    hence

    h(t)=2(0)−1=−1h(t)=2(0)-1=-1h(t)=2(0)−1=−1

    Case 2: t→0−t\to 0^-t→0−

    For −1<t<0-1<t<0−1<t<0,

    [t]=−1,f(t)=−1[t]=-1, \qquad f(t)=-1[t]=−1,f(t)=−1

    hence

    h(t)=2(−1)−(−1)=−2+1=−1h(t)=2(-1)-(-1)=-2+1=-1h(t)=2(−1)−(−1)=−2+1=−1

    Also at t=0t=0t=0,

    [0]=0,f(0)=1  ⟹  h(0)=2(0)−1=−1[0]=0, \quad f(0)=1 \implies h(0)=2(0)-1=-1[0]=0,f(0)=1⟹h(0)=2(0)−1=−1

    Therefore, for all ttt sufficiently close to 000, we get

    h(t)=−1h(t)=-1h(t)=−1

  4. Now evaluate g(h(t))g(h(t))g(h(t))

    Since h(t)=−1h(t)=-1h(t)=−1 near t=0t=0t=0,

    g(h(t))=g(−1)g(h(t))=g(-1)g(h(t))=g(−1)

    By definition of ggg,

    g(−1)=1g(-1)=1g(−1)=1

    Therefore,

    lim⁡t→0g(h(t))=1\lim_{t\to 0} g(h(t))=1t→0lim​g(h(t))=1

  5. Final answer

    1\boxed{1}1​

    So the correct option is A.

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