Question

What happens if you try to use l’Hospital’s Rule to find the limit? Evaluate the limit using another method.

77. \(\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + 1} }}\)

Short answer

The repeated applications of l’Hospital’s Rule gives the original limit

\(\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + 1} }} = 1\).

Step-by-step solution

Write l’Hospital’s Rule

l’Hospital’s Rule:Assume \(f\) and \(g\) are differentiable and \(g'\left( x \right) \ne 0\) then\(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\).

Step 2: What happens if you try to use l’Hospital’s Rule to find the limit

Consider the limit \(\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + 1} }}\).

Apply the l’Hospital’s Rule.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + 1} }} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left( x \right)}}{{\frac{d}{{dx}}\left( {\sqrt {{x^2} + 1} } \right)}}\\ = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{\frac{1}{2}{{\left( {{x^2} + 1} \right)}^{ - \frac{1}{2}}}\left( {2x} \right)}}\\ = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} + 1} }}{x}\end{array}\)

As repeated applications of l’Hospital’s Rule gives the original limit.

Step 3: Solve the limit by another method

Consider the limit \(\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + 1} }}\).

Divide the numerator and denominator of the limit by \(x\).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + 1} }} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{x}{x}}}{{\sqrt {\frac{{{x^2}}}{{{x^2}}} + \frac{1}{{{x^2}}}} }}\\ = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{\sqrt {1 + \frac{1}{{{x^2}}}} }}\\ = \frac{1}{1}\end{array}\)

Thus, \(\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + 1} }} = 1\)