Question

3 evaluate the following limit.
(a) ∞
(b) 2
(c) 1
(d) e
(e) ( e ^ { 2 } )
( lim _ { x
ightarrow infty } left( e ^ { 2 x } + x
ight) ^ { \frac { 1 } { x } } )
( \begin{array} { l } { ln y = \frac { 1 } { x } ln left( e ^ { 2 x } + x
ight) } \\ { = \frac { ln left( e ^ { 2 x } + x
ight) } { x } } \\ { = \frac { \frac { 1 } { left( e ^ { 2 x } + x
ight) } left( 2 e ^ { 2 x } + 1
ight) } { 1 } } \\ { = \frac { 2 e ^ { 2 x } + 1 } { e ^ { 2 x } + 1 ) } } \\ { = \frac { 4 e ^ { 2 x } } { 2 e ^ { 2 x } } } \\ { = 2 } \\ { e ^ { 2 } } end{array} )

Explanation:

Step1: Let \( y=(e^{2x}+x)^{\frac{1}{x}} \)

Take the natural logarithm of both sides: \( \ln y=\frac{\ln(e^{2x}+x)}{x} \)

Step2: Apply L - H rule (since \( \lim_{x

ightarrow\infty}\frac{\ln(e^{2x}+x)}{x} \) is of the form \( \frac{\infty}{\infty} \))
Differentiate the numerator and denominator. The derivative of \( \ln(e^{2x}+x) \) is \( \frac{2e^{2x} + 1}{e^{2x}+x} \), and the derivative of \( x \) is \( 1 \). So, \( \lim_{x
ightarrow\infty}\ln y=\lim_{x
ightarrow\infty}\frac{2e^{2x}+1}{e^{2x}+x} \)

Step3: Simplify the limit

Divide numerator and denominator by \( e^{2x} \): \( \lim_{x
ightarrow\infty}\frac{2+\frac{1}{e^{2x}}}{1+\frac{x}{e^{2x}}} \). Since \( \lim_{x
ightarrow\infty}\frac{x}{e^{2x}} = 0 \) (using L - H rule again: \( \lim_{x
ightarrow\infty}\frac{x}{e^{2x}}=\lim_{x
ightarrow\infty}\frac{1}{2e^{2x}} = 0 \)) and \( \lim_{x
ightarrow\infty}\frac{1}{e^{2x}}=0 \), we get \( \lim_{x
ightarrow\infty}\ln y = 2 \)

Step4: Find the original limit

Since \( y = e^{\ln y} \), then \( \lim_{x
ightarrow\infty}y=e^{\lim_{x
ightarrow\infty}\ln y} \). Substitute \( \lim_{x
ightarrow\infty}\ln y = 2 \), so \( \lim_{x
ightarrow\infty}(e^{2x}+x)^{\frac{1}{x}}=e^{2} \)

Answer:

E. \( e^{2} \)