A) This particular limit is of the indeterminate form,
![\frac{ \infty }{ \infty }](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Cinfty%20%7D%7B%20%5Cinfty%20%7D%20)
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
![Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_ {x \rightarrow \infty } \frac{( ln(x ^{2} + 1 ) ) '}{x ' }](https://tex.z-dn.net/?f=%20Lim_%7Bx%20%5Crightarrow%20%5Cinfty%20%7D%20%5Cfrac%7B%20ln%28x%20%5E%7B2%7D%20%2B%201%20%29%20%7D%7Bx%7D%20%3D%20Lim_%20%7Bx%20%5Crightarrow%20%5Cinfty%20%7D%20%5Cfrac%7B%28%20ln%28x%20%5E%7B2%7D%20%2B%201%20%29%20%29%20%27%7D%7Bx%20%27%20%7D%20)
So we take the derivatives and obtain,
![Lim_ {x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ \frac{2x}{x^{2} + 1} }{1}](https://tex.z-dn.net/?f=%20Lim_%20%7Bx%20%5Crightarrow%20%5Cinfty%20%7D%20%5Cfrac%7B%20ln%28x%20%5E%7B2%7D%20%2B%201%20%29%20%7D%7Bx%7D%20%3D%20Lim_%7Bx%20%5Crightarrow%20%5Cinfty%20%7D%20%5Cfrac%7B%20%5Cfrac%7B2x%7D%7Bx%5E%7B2%7D%20%2B%201%7D%20%7D%7B1%7D%20)
Still it is of the same indeterminate form, so we apply the rule again,
![Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 2 }{2x}](https://tex.z-dn.net/?f=%20Lim_%7Bx%20%5Crightarrow%20%5Cinfty%20%7D%20%5Cfrac%7B%20ln%28x%20%5E%7B2%7D%20%2B%201%20%29%20%7D%7Bx%7D%20%3D%20Lim_%7Bx%20%5Crightarrow%20%5Cinfty%20%7D%20%5Cfrac%7B%202%20%7D%7B2x%7D%20)
This simplifies to,
![Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 1 }{x} = 0](https://tex.z-dn.net/?f=%20Lim_%7Bx%20%5Crightarrow%20%5Cinfty%20%7D%20%5Cfrac%7B%20ln%28x%20%5E%7B2%7D%20%2B%201%20%29%20%7D%7Bx%7D%20%3D%20Lim_%7Bx%20%5Crightarrow%20%5Cinfty%20%7D%20%5Cfrac%7B%201%20%7D%7Bx%7D%20%3D%200%20)
b) This limit is also of the indeterminate form,
![\frac{0}{0}](https://tex.z-dn.net/?f=%20%5Cfrac%7B0%7D%7B0%7D%20)
we still apply the L'Hopital's Rule,
![Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ (tanx)'}{x ' }](https://tex.z-dn.net/?f=%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%5Cfrac%7B%20tanx%7D%7Bx%7D%20%3D%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%20%5Cfrac%7B%20%28tanx%29%27%7D%7Bx%20%27%20%7D%20)
![Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (x) }{1 }](https://tex.z-dn.net/?f=%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%5Cfrac%7B%20tanx%7D%7Bx%7D%20%3D%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%20%5Cfrac%7B%20%5Csec%20%5E%7B2%7D%20%28x%29%20%7D%7B1%20%7D%20)
When we plug in zero now we obtain,
![Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (0) }{1 } = \frac{1}{1} = 1](https://tex.z-dn.net/?f=Lim_%20%7Bx%20%5Crightarrow0%20%7D%5Cfrac%7B%20tanx%7D%7Bx%7D%20%3D%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%20%5Cfrac%7B%20%5Csec%20%5E%7B2%7D%20%280%29%20%7D%7B1%20%7D%20%3D%20%5Cfrac%7B1%7D%7B1%7D%20%3D%201)
c) This also in the same indeterminate form
![Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ ({e}^{2x} - 1 - 2x)'}{( {x}^{2} ) ' }](https://tex.z-dn.net/?f=%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%5Cfrac%7B%20%7Be%7D%5E%7B2x%7D%20-%201%20-%202x%7D%7B%20%7Bx%7D%5E%7B2%7D%20%7D%20%3D%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%20%5Cfrac%7B%20%28%7Be%7D%5E%7B2x%7D%20-%201%20-%202x%29%27%7D%7B%28%20%7Bx%7D%5E%7B2%7D%20%29%20%27%20%7D%20)
![Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (2{e}^{2x} - 2)}{ 2x }](https://tex.z-dn.net/?f=%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%5Cfrac%7B%20%7Be%7D%5E%7B2x%7D%20-%201%20-%202x%7D%7B%20%7Bx%7D%5E%7B2%7D%20%7D%20%3D%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%20%5Cfrac%7B%20%282%7Be%7D%5E%7B2x%7D%20-%202%29%7D%7B%202x%20%7D%20)
It is still of that indeterminate form so we apply the rule again, to obtain;
![Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (4{e}^{2x} )}{ 2 }](https://tex.z-dn.net/?f=%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%5Cfrac%7B%20%7Be%7D%5E%7B2x%7D%20-%201%20-%202x%7D%7B%20%7Bx%7D%5E%7B2%7D%20%7D%20%3D%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%20%5Cfrac%7B%20%284%7Be%7D%5E%7B2x%7D%20%29%7D%7B%202%20%7D%20)
Now we have remove the discontinuity, we can evaluate the limit now, plugging in zero to obtain;
![Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = \frac{ (4{e}^{2(0)} )}{ 2 }](https://tex.z-dn.net/?f=%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%5Cfrac%7B%20%7Be%7D%5E%7B2x%7D%20-%201%20-%202x%7D%7B%20%7Bx%7D%5E%7B2%7D%20%7D%20%3D%20%5Cfrac%7B%20%284%7Be%7D%5E%7B2%280%29%7D%20%29%7D%7B%202%20%7D%20)
This gives us;
![Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } =\frac{ (4(1) )}{ 2 }=2](https://tex.z-dn.net/?f=%20Lim_%20%7Bx%20%5Crightarrow0%20%7D%5Cfrac%7B%20%7Be%7D%5E%7B2x%7D%20-%201%20-%202x%7D%7B%20%7Bx%7D%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B%20%284%281%29%20%29%7D%7B%202%20%7D%3D2%20)
d)
![Lim_ {x \rightarrow +\infty }\sqrt{x^2+2x}-x](https://tex.z-dn.net/?f=%20Lim_%20%7Bx%20%5Crightarrow%20%2B%5Cinfty%20%7D%5Csqrt%7Bx%5E2%2B2x%7D-x)
For this kind of question we need to rationalize the radical function, to obtain;
![Lim_ {x \rightarrow +\infty }\frac{2x}{\sqrt{x^2+2x}+x}](https://tex.z-dn.net/?f=%20Lim_%20%7Bx%20%5Crightarrow%20%2B%5Cinfty%20%7D%5Cfrac%7B2x%7D%7B%5Csqrt%7Bx%5E2%2B2x%7D%2Bx%7D)
We now divide both the numerator and denominator by x, to obtain,
![Lim_ {x \rightarrow +\infty }\frac{2}{\sqrt{1+\frac{2}{x}}+1}](https://tex.z-dn.net/?f=%20Lim_%20%7Bx%20%5Crightarrow%20%2B%5Cinfty%20%7D%5Cfrac%7B2%7D%7B%5Csqrt%7B1%2B%5Cfrac%7B2%7D%7Bx%7D%7D%2B1%7D)
This simplifies to,