The value of limx- x - 6x - 1 x - 4 is

The correct option is D e^5 lim_x→∞ (x + 6x + 1 )^x + 4 #160; = lim_x→∞ (1 + 5x + 1 )^x + 4 = lim_x→∞ (1 + 1x + 1/5 )^x + 4 = lim_x→ ...

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Byju's Answer

Standard XII

Mathematics

Derivative from First Principle

The value of ...

Question

The value of $lim x \to \infty {(\frac{x + 6}{x + 1})}^{x + 4}$ is

e

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e^{2}

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e^{4}

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e^{5}

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Solution

The correct option is D $e^{5}$
$lim x \to \infty {(\frac{x + 6}{x + 1})}^{x + 4}$
$= lim x \to \infty {(1 + \frac{5}{x + 1})}^{x + 4}$
$= lim x \to \infty {⎛ ⎝ 1 + \frac{1}{\frac{x + 1}{5}} ⎞ ⎠}^{x + 4}$
$= lim x \to \infty {⎡ ⎣ {(1 + \frac{5}{x + 1})}^{\frac{x + 1}{5}} ⎤ ⎦}^{lim x \to \infty \frac{5}{x + 1} \times (x + 4)}$
$= e^{lim x \to \infty \frac{5 (x + 4)}{(x + 1)}}$
$= e^{5}$

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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 =____________________.

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?

Q. If

E = E_{1} \cap E_{2} \cap E_{3} \cap E_{4} \cap E_{5}

and

P (E_{1}) = \frac{95}{100}, P (E_{2} / E_{1}) = \frac{94}{99}, P (E_{3} / (E_{1} \cap E_{2})) = \frac{93}{98}, P (E_{4} / E_{1} \cap E_{2} \cap E_{3}) = \frac{92}{97}

and

P (E_{5} / E_{1} \cap E_{2} \cap E_{3} \cap E_{4}) = \frac{91}{96}

, then

P (E)

is:

Q. If

E = E_{1} E_{2} E_{3} E_{4} E_{5}

and

P (E_{1}) = \frac{95}{100}

P (E_{2} / E_{1}) = \frac{94}{99}

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P (E_{4} / E_{1} E_{2} E_{3}) = \frac{92}{97}

and

P (E_{5} / E_{1} E_{2} E_{3} E_{4}) = \frac{91}{96}

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Q. The value of the limit

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The value of limx→∞(x+6x+1)x+4 is

The correct option is D e5limx→∞(x+6x+1)x+4 =limx→∞(1+5x+1)x+4=limx→∞⎛⎝1+1x+15⎞⎠x+4=limx→∞⎡⎣(1+5x+1)x+15⎤⎦limx→∞5x+1×(x+4)=elimx→∞5(x+4)(x+1) =e5

The value of $lim x \to \infty {(\frac{x + 6}{x + 1})}^{x + 4}$ is