Question

what happens if you try to use lhospitals rule to find the limit?
\\( \lim _ { x \
ightarrow \infty } \frac { x } { \sqrt { x ^ { 2 } + 4 } } \\)
you cannot apply lhospitals rule because the function is not continuous.
you cannot apply lhospitals rule because the function is not differentiable.
you cannot apply lhospitals rule because the denominator equals zero for some value \\( x = \\)
a.
you cannot apply lhospitals rule because the numerator equals zero for some value \\( x = a \\).
repeated applications of lhospitals rule result in the original limit or the limit of thereciprocal of the function.
evaluate the limit using another method.

Explanation:

Step1: Apply L'Hospital's Rule

Differentiate numerator and denominator.
The derivative of \(y = x\) is \(y'=1\).
The derivative of \(y=\sqrt{x^{2}+4}=(x^{2} + 4)^{\frac{1}{2}}\) using the chain - rule \((u^{n})'=nu^{n - 1}u'\) where \(u=x^{2}+4\), \(n=\frac{1}{2}\).
\(y'=\frac{1}{2}(x^{2}+4)^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{x^{2}+4}}\)
So, \(\lim_{x
ightarrow\infty}\frac{x}{\sqrt{x^{2}+4}}\) becomes \(\lim_{x
ightarrow\infty}\frac{1}{\frac{x}{\sqrt{x^{2}+4}}}=\lim_{x
ightarrow\infty}\frac{\sqrt{x^{2}+4}}{x}\)

Step2: Simplify the original limit without L'Hospital's Rule

Divide numerator and denominator by \(x\) (since \(x
ightarrow\infty,x>0\)).
\(\lim_{x
ightarrow\infty}\frac{x}{\sqrt{x^{2}+4}}=\lim_{x
ightarrow\infty}\frac{x}{|x|\sqrt{1 + \frac{4}{x^{2}}}}\)
Since \(x
ightarrow\infty,x>0\), \(|x| = x\).
\(\lim_{x
ightarrow\infty}\frac{x}{x\sqrt{1+\frac{4}{x^{2}}}}=\lim_{x
ightarrow\infty}\frac{1}{\sqrt{1+\frac{4}{x^{2}}}}\)

Step3: Evaluate the limit

As \(x
ightarrow\infty,\frac{4}{x^{2}}
ightarrow0\).
\(\lim_{x
ightarrow\infty}\frac{1}{\sqrt{1+\frac{4}{x^{2}}}} = 1\)

Answer:

Repeated applications of l'Hospital's Rule result in the original limit or the limit of the reciprocal of the function. The value of the limit \(\lim_{x
ightarrow\infty}\frac{x}{\sqrt{x^{2}+4}}\) is \(1\)