L'Hopital's rule does not help with the given limit. Find the limit some other way.
-$\lim _ { x \rightarrow\infty } \frac { \sqrt { 25 x + 1 } } { \sqrt { x + 1 } }$
A) 25
B) $\infty$
C) 0
D) 5

Question

L'Hopital's rule does not help with the given limit. Find the limit some other way.
-$\lim _ { x \rightarrow\infty  } \frac { \sqrt { 25 x + 1 } } { \sqrt { x + 1 } }$
A) 25
B) $\infty$
C) 0
D) 5

Quizplus · Accepted Answer

We can use the fact that as x goes to infinity, the higher degree terms dominate the denominator. Therefore, we can simplify the fraction by dividing both numerator and denominator by x:
$$\lim_{x\rightarrow\infty}\frac{\sqrt{25x+1}}{\sqrt{x+1}}=\lim_{x\rightarrow\infty}\frac{\frac{\sqrt{25x+1}}{x}}{\frac{\sqrt{x+1}}{x}}$$
Now, we can use the fact that as x goes to infinity, both the numerator and denominator go to 0/0 form. We can apply L'Hopital's rule once to get:
$$\lim_{x\rightarrow\infty}\frac{\frac{1}{2\sqrt{25x+1}}\cdot 25}{\frac{1}{2\sqrt{x+1}}}$$
Simplifying, we get:
$$\lim_{x\rightarrow\infty}\frac{25\sqrt{x+1}}{\sqrt{25x+1}}=\lim_{x\rightarrow\infty}\frac{25}{\sqrt{1+1/x}}=25$$
Therefore, the answer is D.

L'Hopital's Rule Does Not Help with the Given Limit lim⁡x→∞25x+1x+1\lim _ { x \rightarrow\infty } \frac { \sqrt { 25 x + 1 } } { \sqrt { x + 1 } }limx→∞​x+1​25x+1​​

L'Hopital's Rule Does Not Help with the Given Limit $\lim _ { x \rightarrow\infty } \frac { \sqrt { 25 x + 1 } } { \sqrt { x + 1 } }$