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Limits — JEE Maths practice question

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

Question

Let

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)=\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?

\displaystyle\lim_{x\to 0^{-}}g(f(x))=5

correct

\displaystyle\lim_{x\to 0^{-}}g(f(x))=7

\displaystyle\lim_{x\to 0^{+}}g(f(x))=5

correct

\displaystyle\lim_{x\to 0^{+}}g(f(x))=-1

Solution

Step 1: Compute

\displaystyle\lim_{x\to 0^{-}}f(x)

. For

x<0

f(x)=\dfrac{x}{\tan x}

. Let

h=-x\to 0^{+}

\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\to 0^{-}

f(x)\to 1

, but slightly less than

1

(since

\dfrac{h}{\tan h}<1

for small

h>0

). Step 2: Compute

\displaystyle\lim_{x\to 0^{+}}f(x)

. For

x>0

f(x)=\dfrac{\ln(1+2x)}{x}=\dfrac{\ln(1+2x)}{2x}\cdot 2\to 1\cdot 2=2

. So as

x\to 0^{+}

f(x)\to 2

, slightly less than

2

. Step 3: Compute

\displaystyle\lim_{x\to 0^{-}}g(f(x))

. As

x\to 0^{-}

f(x)\to 1^{-}

(just below

1

). Use the first piece of

g

(since input

<1

g(f(x))=f(x)+4\to 1+4=5

Option (1) is correct. Step 4: Compute

\displaystyle\lim_{x\to 0^{+}}g(f(x))

. As

x\to 0^{+}

f(x)\to 2^{-}

(just below

2

). Use the middle piece of

g

(since input is in

[1,2)

g(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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