First, substitute in \(2\) to see whether we arrive at an indeterminate form.

\[\mathop {\lim }\limits_{x \to 4}\frac{\sin\left(\pi x\right)}{x^{2}-5x+6}=\frac{\sin\left(2\pi\right)}{2^{2}-10+6}=\frac{0}{0}\]

Since \(\frac{0}{0}\) is an indeterminate form, we can use L'Hôpital's rule.

\[\mathop {\lim }\limits_{x \to 2}\frac{\sin\left(\pi x\right)}{x^{2}-5x+6}=\mathop {\lim }\limits_{x \to 2} \frac{\frac{d}{dx}\left[\sin\left(\pi x\right)\right]}{\frac{d}{dx}\left[x^{2}-5x+6\right]}=\]

\[\mathop {\lim }\limits_{x \to 2} \frac{\pi\cos\left(\pi x\right)}{2x-5}=\frac{\pi\cos\left(\pi x\right)}{2x-5}=\frac{\pi\cos\left(2\pi\right)}{4-5}=\]\[\frac{\pi\left(1\right)}{-1}=-\pi\]

We were already able to solve these types of limits by dividing the numerator and the denominator by the highest power term in the denominator; however, we can also solve them by using L'Hôpital's Rule.

First, check to see whether direct substitution yields an indeterminate form.

\[\mathop {\lim }\limits_{x \to 2} \frac{2x^{3}+2x^{2}\ +2}{8x^{2}+9x^{3}+4x}=\frac{\infty}{\infty}\] This is true because both the numerator and the denominator will tend towards \(\infty\). \(\frac{\infty}{\infty}\) is an indeterminate form; therefore, we can use L'Hôpital's Rule.

\[\mathop {\lim }\limits_{x \to 2} \frac{2x^{3}+2x^{2}\ +2}{8x^{2}+9x^{3}+4x}=\mathop {\lim }\limits_{x \to 2}\frac{\frac{d}{dx}\left[2x^{3}+2x^{2}\ +2\right]}{\frac{d}{dx}\left[8x^{2}+9x^{3}+4x\right]}=\] \[\mathop {\lim }\limits_{x \to 2} \frac{6x^{2}+4x}{16x+27x^{2}}=\frac{\infty}{\infty}\]

Since we got an indeterminate form again, we can use L'Hôpital's Rule again.

\[\mathop {\lim }\limits_{x \to 2} \frac{6x^{2}+4x}{16x+27x^{2}}= \mathop {\lim }\limits_{x \to 2} \frac{\frac{d}{dx}\left[6x^{2}+4x\right]}{\frac{d}{dx}\left[16x+27x^{2}\right]}=\mathop {\lim }\limits_{x \to 2} \frac{12x+4}{16+54x}=\frac{\infty}{\infty}\]

Since we got an indeterminate form again, we will use L'Hôpital's Rule again.

\[\mathop {\lim }\limits_{x \to 2} \frac{12x+4}{16+54x}= \mathop {\lim }\limits_{x \to 2}\frac{\frac{d}{dx}\left[12x+4\right]}{\frac{d}{dx}\left[16+54x\right]}=\] \[\mathop {\lim }\limits_{x \to 2}\frac{12}{54}=\frac{2\left(6\right)}{9\left(6\right)}=\frac{2}{9}\]

\[\therefore\]

\[\mathop {\lim }\limits_{x \to 2} \frac{2x^{3}+2x^{2}\ +2}{8x^{2}+9x^{3}+4x}=\frac{2}{9}\]

First, check to see whether direct substitution yields an indeterminate form.

\[\mathop {\lim }\limits_{x \to 5} \frac{\sqrt{x+20}-5}{x-5}=\frac{\sqrt{25}-5}{5-5}=\frac{0}{0}\]

Since we got an indeterminate form again, we will use L'Hôpital's Rule.

\[\mathop {\lim }\limits_{x \to 5} \frac{\sqrt{x+20}-5}{x-5}=\mathop {\lim }\limits_{x \to 5}\frac{\frac{d}{dx}\left[\sqrt{x+20}-5\right]}{\frac{d}{dx}\left[x-5\right]}=\mathop {\lim }\limits_{x \to 5}\frac{\frac{d}{dx}\left[\left(x+20\right)^{\frac{1}{2}}-5\right]}{1}=\] \[\mathop {\lim }\limits_{x \to 5}\frac{1}{2}\left(x+20\right)^{-\frac{1}{2}}=\mathop {\lim }\limits_{x \to 5}\frac{1}{2\sqrt{x+20}}=\frac{1}{10}\]

\[\therefore\]

\[\mathop {\lim }\limits_{x \to 5} \frac{\sqrt{x+20}-5}{x-5}=\frac{1}{10}\]

First, check if evaluating the limit as it is, yields an indeterminate form.

If we approach \(1\) from the right, both fractions in the limit tend towards \(\infty\) since the denominator gets infinitely small while the numerator stays constant.

\[\therefore\]

\[\mathop {\lim }\limits_{x \to 1^+} \frac{3}{x-2}-\frac{2}{\ln\left(x-1\right)}=\infty-\infty\]

Since we \(\infty-\infty\) is an indeterminate, we can use L'Hôpital's Rule; however, we need to manipulate it first since the indeterminate form is neither \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

To do this we will combine the two terms by finding a common denominator.

\[\mathop {\lim }\limits_{x \to 1^+}\frac{3}{x-2}-\frac{2}{\ln\left(x-1\right)}=\mathop {\lim }\limits_{x \to 1^+}\frac{3\ln\left(x-1\right)-2\left(x-2\right)}{\left(x-2\right)\left(\ln\left(x-1\right)\right)}=\] \[\mathop {\lim }\limits_{x \to 1^+}\frac{3\ln\left(x-1\right)-2x+4}{\left(x-2\right)\left(\ln\left(x-1\right)\right)}\]

If we solve the above limit, we will get \(\frac{0}{0}\); therefore, we can use L'Hôpital's Rule.

\[\mathop {\lim }\limits_{x \to 1^+}\frac{3\ln\left(x-1\right)-2x+4}{\left(x-2\right)\left(\ln\left(x-1\right)\right)}=\mathop {\lim }\limits_{x \to 1^+}\frac{\frac{d}{dx}\left[3\ln\left(x-1\right)-2x+4\right]}{\frac{d}{dx}\left[\left(x-2\right)\left(\ln\left(x-1\right)\right)\right]}\] \[\mathop {\lim }\limits_{x \to 1^+}\frac{3\frac{d}{dx}\left[\ln\left(x-1\right)\right]-2}{\frac{d}{dx}\left[\left(x-2\right)\right]\left(\ln\left(x-1\right)\right)+\frac{d}{dx}\left[\ln\left(x-1\right)\right]\left(x-2\right)}=\] \[\mathop {\lim }\limits_{x \to 1^+}\frac{\frac{3}{x-1}-2}{\ln\left(x-1\right)+\frac{1}{x-1}\left(x-2\right)}=\mathop {\lim }\limits_{x \to 1^+}\frac{\frac{3}{x-1}-2}{\ln\left(x-1\right)+\frac{x-2}{x-1}}\]

Now, let's evaluate the above limit. If we approach \(1\) from the right the numerator will tend towards \(\infty\), while the denominator will tend towards \(0\); therefore, the limit will tend towards \(\infty\) as \(x\) approaches \(1\).

\[\mathop {\lim }\limits_{x \to 1^+}\frac{\frac{3}{x-1}-2}{\ln\left(x-1\right)+\frac{x-2}{x-1}}=\infty\]

First, check if evaluating the limit as it is, yields an indeterminate form.

As we approach \(\infty\), \(\left(e^{x}+x\right)\) tends towards \(\infty\) while \(\frac{1}{x}\) tends towards as \(0\); therefore, the limit is equal to \({\infty}^{0}\); which is an indeterminate form. We can use L'Hôpital's rule if we manipulate the function. Since \({\infty}^{0}\) is an indeterminate power we will take the natural log of both sides of the function so that the limit yields either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

\[y=\left(e^{x}+x\right)^{\frac{1}{x}}\]

\[\ln\left(y\right)=\ln\left(\left(e^{x}+x\right)^{\frac{1}{x}}\right)=\frac{1}{x}\ln\left(e^{x}+x\right)=\] \[\frac{\ln\left(e^{x}+x\right)}{x}\]

Now evaluate the limit using L'Hôpital's rule.

\[\mathop {\lim }\limits_{x \to \infty}\ln\left(y\right)=\mathop {\lim }\limits_{x \to \infty}=\frac{\ln\left(e^{x}+x\right)}{x}=\] \[\mathop {\lim }\limits_{x \to \infty}\frac{\frac{d}{dx}\left[\ln\left(e^{x}+x\right)\right]}{\frac{d}{dx}\left[x\right]}=\mathop {\lim }\limits_{x \to \infty}\frac{\frac{1}{e^{x}+x}\cdot\frac{d}{dx}\left[e^{x}+x\right]}{1}=\] \[\mathop {\lim }\limits_{x \to \infty}\frac{e^{x}+1}{e^{x}+x}=\frac{\infty}{\infty}\]

Since the limit is equal to the indeterminate form \(\frac{\infty}{\infty}\), we will use L'Hôpital's rule again.

\[\mathop {\lim }\limits_{x \to \infty}\ln\left(y\right)=\mathop {\lim }\limits_{x \to \infty}\frac{e^{x}+1}{e^{x}+x}=\] \[\mathop {\lim }\limits_{x \to \infty}\frac{\frac{d}{dx}\left[e^{x}+1\right]}{\frac{d}{dx}\left[e^{x}+x\right]}=\] \[\mathop {\lim }\limits_{x \to \infty}\frac{e^{x}}{e^{x}+1}=\frac{\infty}{\infty}\]

Since we get the indeterminate form \(\frac{\infty}{\infty}\) again, we will use L'Hôpital's rule once more.

\[\mathop {\lim }\limits_{x \to \infty}\ln\left(y\right)=\mathop {\lim }\limits_{x \to \infty}\frac{e^{x}}{e^{x}+1}=\mathop {\lim }\limits_{x \to \infty}\frac{\frac{d}{dx}\left[e^{x}\right]}{\frac{d}{dx}\left[e^{x}+1\right]}=\] \[\mathop {\lim }\limits_{x \to \infty}\frac{e^{x}}{e^{x}}=\mathop {\lim }\limits_{x \to \infty}1=1\]

Since we got an answer that is not an indeterminate form, we no longer have to use L'Hôpital's rule; however, the limit we calculuated was for \[\ln\left(y\right)\] and were asked to compute the limit for \(y\); therefore, we need to solve for the function \(y\).

\[\mathop {\lim }\limits_{x \to \infty}\ln\left(y\right)=1\]

\[\mathop {\lim }\limits_{x \to \infty}e^{\ln\left(y\right)}=e^{1}\]

\[\mathop {\lim }\limits_{x \to \infty}y=e\]

\[\therefore\]

\[\mathop {\lim }\limits_{x \to \infty}\left(e^{x}+x\right)^{\frac{1}{x}}=e\]

First, check if evaluating the limit as it is, yields an indeterminate form.

As we approach \(0\) from the right, both \(\tan\left(3x\right)\) and \(x\) will approach \(0\); therefore, the limit will be equal to the indeterminate form \(0^{0}\). Since \(0^{0}\) is an indeterminate power, we can algebraically manipulate it so that we can use L'Hôpital's rule.

First, take the natural log of both sides of the function and evaluate the limit.

\[\ln\left(y\right)=\ln\left(\left(\tan\left(3x\right)\right)^{x}\right)=x\ln\left(\tan\left(3x\right)\right)\]

\[\mathop {\lim }\limits_{x \to 0^+} \ln\left(y\right) = \mathop {\lim }\limits_{x \to 0^+} x\ln\left(\tan\left(3x\right)\right)\]

As \(x\) approaches \(0\) from the right, \(x\) will approach \(0\) while \(\ln\left(\tan\left(3x\right)\right)\) will tend towards \(-\infty\) (since \(\tan\left(3x\right)\) approaches \(0\), \(\ln\left(\tan\left(3x\right)\right)\) will tend towards \(-\infty\)); therefore, the limit is equal to \(-0\cdot\infty\), which is the indeterminate product. Since we get an indeterminate product, we have to algebraically manipulate the function once more using the formulas so that we can use L'Hôpital's rule.

\[\ln\left(y\right)=x\ln\left(\tan\left(3x\right)\right)=\frac{\ln\left(\tan\left(3x\right)\right)}{\frac{1}{x}}\]

\[\mathop {\lim }\limits_{x \to 0^+} \ln\left(y\right) =\mathop {\lim }\limits_{x \to 0^+} \frac{\ln\left(\tan\left(3x\right)\right)}{\frac{1}{x}}=\mathop {\lim }\limits_{x \to 0^+}\frac{\frac{d}{dx}\left[\ln\left(\tan\left(3x\right)\right)\right]}{\frac{d}{dx}\left[\frac{1}{x}\right]}=\] \[\mathop {\lim }\limits_{x \to 0^+}\frac{\frac{1}{\tan\left(3x\right)}\cdot\frac{d}{dx}\left[\tan\left(3x\right)\right]}{-\frac{1}{x^{2}}}=-\mathop {\lim }\limits_{x \to 0^+}\frac{\frac{3}{\tan\left(3x\right)}\cdot\sec^{2}\left(3x\right)}{\frac{1}{x^{2}}}\]

Before we evaluate the above imit, let's simplify it.

\[\mathop {\lim }\limits_{x \to 0^+} \ln\left(y\right) =-\mathop {\lim }\limits_{x \to 0^+}\left[\frac{3}{\tan\left(3x\right)}\cdot\sec^{2}\left(3x\right)\right]x^{2}=\]

\[-\mathop {\lim }\limits_{x \to 0^+}\left[\frac{3}{\frac{\sin\left(3x\right)}{\cos\left(3x\right)}}\cdot\frac{1}{\cos^{2}\left(3x\right)}\right]x^{2}=-\mathop {\lim }\limits_{x \to 0^+}\left[\frac{3x^{2}}{\sin\left(3x\right)}\cdot\frac{1}{\cos\left(3x\right)}\right]=\] \[-\mathop {\lim }\limits_{x \to 0^+}\left[\frac{3x^{2}}{\sin\left(3x\right)}\cdot\sec\left(3x\right)\right]\]

Let's now divide the limit into two parts using the properties of limits.

\[\mathop {\lim }\limits_{x \to 0^+} \ln\left(y\right)=-\left[[\mathop {\lim }\limits_{x \to 0^+}\frac{3x^{2}}{\sin\left(3x\right)}\cdot\mathop {\lim }\limits_{x \to 0^+}\sec\left(3x\right)\right]\]

Let's try and evaluate the above limit starting with the second function in the product. As \(x\) approaches \(0\) from the right, the secant function will approach \(1\). As \(x\) approaches \(0\) from the right both \(3x^{2}\) and \(\sin\left(3x\right)\) will approach \(0\); therefore, will get the indeterminate form \(\frac{0}{0}\) and we will need to evaluate the first limit again using L'Hôpital's Rule.

\[\mathop {\lim }\limits_{x \to 0^+} \ln\left(y\right)=-1\cdot\mathop {\lim }\limits_{x \to 0^+}\frac{\frac{d}{dx}\left[3x^{2}\right]}{\frac{d}{dx}\left[\sin\left(3x\right)\right]}=\] \[-1\cdot\mathop {\lim }\limits_{x \to 0^+}\frac{6x}{3\cos\left(3x\right)}=-1\cdot\frac{0}{1}=0\]

Now that we finally have an answer that is not an indeterminate form, we need to find \(\mathop {\lim }\limits_{x \to 0^+}y\) (remember that so far we have only found that \(\mathop {\lim }\limits_{x \to 0^+}\ln\left(y\right)=0\)

\[\ln\left(y\right)=0\] \[e^{\ln\left(y\right)}=e^{0}\] \[y=1\] \[\therefore\]

\[\mathop {\lim }\limits_{x \to 0^+}\left(\tan\left(3x\right)\right)^{x}=1\]