Evaluate - lim x - x - 5 x - 2 x - 3 -

The correct option is C e ^ 3 y= x→∞ #160; lim ( x+5 x+2 #160; ) #160; ^ x+3 lny= x→∞ #160; lim ln( x+5 x+2 #160; ) ln( x+3 ) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

Evaluate : ...

Question

Evaluate : ${lim}_{x \to \infty} {(\frac{x + 5}{x + 2})}^{x + 3} = ?$

e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

e^{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $e^{3}$
$y = l i m x \to \infty {(\frac{x + 5}{x + 2})}^{x + 3}$
$l n y = l i m x \to \infty l n (\frac{x + 5}{x + 2}) l n (x + 3) = l i m x \to \infty (x + 3) l n (\frac{1 + 5 / x}{1 + 2 / x})$
$= l i m x \to \infty \frac{l n (\frac{x + 5}{x + 2})}{\frac{1}{(x + 3)}}$ $= l i m x \to \infty \frac{\frac{(x + 2)}{(x + 5)} \times (\frac{(x + 2) - (x + 5)}{{(x + 2)}^{2}})}{\frac{- 1}{{(x + 3)}^{2}}}$
$= - 3 l i m x \to \infty \frac{\frac{1}{x^{2} (1 + 5 / x) (1 + 2 / x)}}{\frac{- 1}{x^{2} {(1 + 3 / x)}^{2}}}$ $= 3 \Rightarrow y = e^{3}$

Suggest Corrections

Similar questions

Q. Evaluate:

\int \frac{e^{6 \log_{e} x} - e^{5 \log_{e} x}}{e^{4 \log_{e} x} - e^{3 \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} E_{2} E_{3} E_{4} E_{5}

and

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

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

P (E_{1} / E_{1} E_{2}) = \frac{93}{98}

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}

, then find

P (E)

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. Evaluate

lim x \to \infty \frac{1}{n} + \frac{e^{\frac{1}{n}}}{n} + \frac{e^{\frac{2}{n}}}{n} + \frac{e^{\frac{3}{n}}}{n} + . . . + \frac{e^{\frac{n - 1}{n}}}{n}

Join BYJU'S Learning Program

Evaluate : limx→∞(x+5x+2)x+3=?

The correct option is C e3y=limx→∞(x+5x+2)x+3lny=limx→∞ln(x+5x+2)ln(x+3)=limx→∞(x+3)ln(1+5/x1+2/x) =limx→∞ln(x+5x+2)1(x+3) =limx→∞(x+2)(x+5)×((x+2)−(x+5)(x+2)2)−1(x+3)2 =−3limx→∞1x2(1+5/x)(1+2/x)−1x2(1+3/x)2 =3⇒y=e3

Evaluate : ${lim}_{x \to \infty} {(\frac{x + 5}{x + 2})}^{x + 3} = ?$