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Limits Continuity and Differentiability question

2025 · 22 Jan · Shift 2 · Q39
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  5. /2025 · 22 Jan · Shift 2 · Q39

Limits Continuity and Differentiability question

2025 · 22 Jan · Shift 2 · Q39

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If lim⁡x→∞((e1−e)(1e−x1+x))x=α\lim_{x \rightarrow \infty}\left(\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\frac{1}{\mathrm{e}}-\frac{x}{1+x}\right)\right)^x=\alphalimx→∞​((1−ee​)(e1​−1+xx​))x=α, then the value of log⁡eα1+log⁡eα\frac{\log _{\mathrm{e}} \alpha}{1+\log _{\mathrm{e}} \alpha}1+loge​αloge​α​ equals :
  1. A
    e−2e^{-2}e−2
  2. B
    e2\mathrm{e}^2e2
  3. C
    eee
  4. D
    e−1e^{-1}e−1
View written solutionFree

Correct answer: C

  1. Given limit

We need to evaluate

α=lim⁡x→∞((e1−e)(1e−x1+x))x.\alpha=\lim_{x\to\infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x.α=x→∞lim​((1−ee​)(e1​−1+xx​))x.

Then find

ln⁡α1+ln⁡α.\frac{\ln \alpha}{1+\ln \alpha}.1+lnαlnα​.
  1. Simplify the base inside the limit

First,

x1+x=1−11+x.\frac{x}{1+x}=1-\frac{1}{1+x}.1+xx​=1−1+x1​.

So,

1e−x1+x=1e−1+11+x.\frac{1}{e}-\frac{x}{1+x} =\frac{1}{e}-1+\frac{1}{1+x}.e1​−1+xx​=e1​−1+1+x1​.

Since

1e−1=1−ee,\frac{1}{e}-1=\frac{1-e}{e},e1​−1=e1−e​,

we get

1e−x1+x=1−ee+11+x.\frac{1}{e}-\frac{x}{1+x}=\frac{1-e}{e}+\frac{1}{1+x}.e1​−1+xx​=e1−e​+1+x1​.

Now multiply by e1−e\dfrac{e}{1-e}1−ee​:

(e1−e)(1−ee+11+x)=1+e(1−e)(1+x).\left(\frac{e}{1-e}\right)\left(\frac{1-e}{e}+\frac{1}{1+x}\right) =1+\frac{e}{(1-e)(1+x)}.(1−ee​)(e1−e​+1+x1​)=1+(1−e)(1+x)e​.

Since 1−e=−(e−1)1-e=-(e-1)1−e=−(e−1),

1+e(1−e)(1+x)=1−e(e−1)(1+x).1+\frac{e}{(1-e)(1+x)}=1-\frac{e}{(e-1)(1+x)}.1+(1−e)(1+x)e​=1−(e−1)(1+x)e​.

Hence,

α=lim⁡x→∞(1−e(e−1)(1+x))x.\alpha=\lim_{x\to\infty}\left(1-\frac{e}{(e-1)(1+x)}\right)^x.α=x→∞lim​(1−(e−1)(1+x)e​)x.
  1. Use the standard exponential limit

Let

c=ee−1.c=\frac{e}{e-1}.c=e−1e​.

Then

α=lim⁡x→∞(1−cx+1)x.\alpha=\lim_{x\to\infty}\left(1-\frac{c}{x+1}\right)^x.α=x→∞lim​(1−x+1c​)x.

Write it as

α=lim⁡x→∞[(1−cx+1)x+1]xx+1.\alpha=\lim_{x\to\infty}\left[\left(1-\frac{c}{x+1}\right)^{x+1}\right]^{\frac{x}{x+1}}.α=x→∞lim​[(1−x+1c​)x+1]x+1x​.

Now,

lim⁡x→∞(1−cx+1)x+1=e−c,\lim_{x\to\infty}\left(1-\frac{c}{x+1}\right)^{x+1}=e^{-c},x→∞lim​(1−x+1c​)x+1=e−c,

and

lim⁡x→∞xx+1=1.\lim_{x\to\infty}\frac{x}{x+1}=1.x→∞lim​x+1x​=1.

Therefore,

α=e−c=e−ee−1.\alpha=e^{-c}=e^{-\frac{e}{e-1}}.α=e−c=e−e−1e​.

So,

ln⁡α=−ee−1.\ln \alpha=-\frac{e}{e-1}.lnα=−e−1e​.
  1. Evaluate the required expression

We need

ln⁡α1+ln⁡α=−ee−11−ee−1.\frac{\ln \alpha}{1+\ln \alpha} =\frac{-\frac{e}{e-1}}{1-\frac{e}{e-1}}.1+lnαlnα​=1−e−1e​−e−1e​​.

Now simplify the denominator:

1−ee−1=e−1−ee−1=−1e−1.1-\frac{e}{e-1}=\frac{e-1-e}{e-1}=\frac{-1}{e-1}.1−e−1e​=e−1e−1−e​=e−1−1​.

Thus,

−ee−1−1e−1=e.\frac{-\frac{e}{e-1}}{-\frac{1}{e-1}}=e.−e−11​−e−1e​​=e.

So the value is

e.\boxed{e}.e​.
  1. Check options
  • A: e−2e^{-2}e−2
  • B: e2e^2e2
  • C: eee
  • D: e−1e^{-1}e−1

Correct option is C.

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