Limx- x - 6x - 1 x - 4 -

The correct option is B e^5 lim_x→∞ (x + 1x + 1 + 5x + 1 )^x + 4 = lim_x→∞ (1 + 5x + 1 )^x + 4 It is in the form of (1 + 1x )^x when ...

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Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

limx→∞ x + 6x...

Question

$lim x \to \infty {(\frac{x + 6}{x + 1})}^{x + 4} =$

e^{4}

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e^{6}

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e^{5}

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e

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Solution

The correct option is B $e^{5}$
$lim x \to \infty {(\frac{x + 1}{x + 1} + \frac{5}{x + 1})}^{x + 4}$ $= lim x \to \infty {(1 + \frac{5}{x + 1})}^{x + 4}$ It is in the form of ${(1 + \frac{1}{x})}^{x}$ when $x \to \infty$ i.e. $lim x \to \infty {(1 + \frac{1}{x})}^{x} = e$ So, $lim x \to \infty {(1 + \frac{5}{x + 1})}^{\frac{(x + 1) \times 5}{5}} {(1 + \frac{5}{x + 1})}^{3}$ $= e^{5} \cdot 1 = e^{5}$

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Similar questions

\int \frac{e^{5 \log_{e} x} - e^{4 \log_{e} x}}{e^{3 \log_{e} x} - e^{2 \log_{e} x}} d x =____________________.

Q. Assertion(A):

lim x \to \infty {(\frac{x^{2} + 5 x + 3}{x^{2} + x + 2})}^{x} = e^{4}

Reason(R):

lim x \to \infty (1 + x)^{1 / x} = e

Q. Solve

lim x \to \infty {(\frac{x + 6}{x + 1})}^{x + 4}

Q. The value of

lim x \to \infty {(\frac{x + 6}{x + 1})}^{x + 4}

The figure gives the electric potential $V$ as a function of distance through five regions on $x -$ axis. Which of the following is true for the electric field $E$ in these regions?

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limx→∞(x+6x+1)x+4=

The correct option is B e5limx→∞(x+1x+1+5x+1)x+4=limx→∞(1+5x+1)x+4It is in the form of (1+1x)x when x→∞ i.e. limx→∞(1+1x)x=eSo, limx→∞(1+5x+1)(x+1)×55(1+5x+1)3=e5⋅1=e5

$lim x \to \infty {(\frac{x + 6}{x + 1})}^{x + 4} =$