a) This particular limit is of the indeterminate form,
[tex] \frac{ \infty }{ \infty } [/tex]
if we plug in infinity directly, though it is not a number just to check.
If a limit is in this form, we apply L'Hopital's Rule.
's
[tex] Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_ {x \rightarrow \infty } \frac{( ln(x ^{2} + 1 ) ) '}{x ' } [/tex]
So we take the derivatives and obtain,
[tex] Lim_ {x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ \frac{2x}{x^{2} + 1} }{1} [/tex]
Still it is of the same indeterminate form, so we apply the rule again,
[tex] Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 2 }{2x} [/tex]
This simplifies to,
[tex] Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 1 }{x} = 0 [/tex]
b) This limit is also of the indeterminate form,
[tex] \frac{0}{0} [/tex]
we still apply the L'Hopital's Rule,
[tex] Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ (tanx)'}{x ' } [/tex]
[tex] Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (x) }{1 } [/tex]
When we plug in zero now we obtain,
[tex]Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (0) }{1 } = \frac{1}{1} = 1[/tex]
c) This also in the same indeterminate form
[tex] Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ ({e}^{2x} - 1 - 2x)'}{( {x}^{2} ) ' } [/tex]
[tex] Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (2{e}^{2x} - 2)}{ 2x } [/tex]
It is still of that indeterminate form so we apply the rule again, to obtain;
[tex] Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (4{e}^{2x} )}{ 2 } [/tex]
Now we have remove the discontinuity, we can evaluate the limit now, plugging in zero to obtain;
[tex] Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = \frac{ (4{e}^{2(0)} )}{ 2 } [/tex]
This gives us;
[tex] Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } =\frac{ (4(1) )}{ 2 }=2 [/tex]
d) [tex] Lim_ {x \rightarrow +\infty }\sqrt{x^2+2x}-x[/tex]
For this kind of question we need to rationalize the radical function, to obtain;
[tex] Lim_ {x \rightarrow +\infty }\frac{2x}{\sqrt{x^2+2x}+x}[/tex]
We now divide both the numerator and denominator by x, to obtain,
[tex] Lim_ {x \rightarrow +\infty }\frac{2}{\sqrt{1+\frac{2}{x}}+1}[/tex]
This simplifies to,
[tex] =\frac{2}{\sqrt{1+0}+1}=1[/tex]
