Solution:

Given limit:

[tex]\displaystyle\tt\longrightarrow L=\lim_{x\to 2}\dfrac{3^{x+2}-81}{9^x-9^2}[/tex]

If we substitute x = 2, we get:

[tex]\displaystyle\tt\longrightarrow L=\dfrac{3^{4}-81}{9^2-9^2}[/tex]

[tex]\displaystyle\tt\longrightarrow L=\dfrac{0}{0}[/tex]

Which is indeterminate form. So, we will modify the expression so as to get the desired result.

[tex]\displaystyle\tt\longrightarrow L=\lim_{x\to 2}\dfrac{3^{x+2}-81}{9^x-9^2}[/tex]

Can be written as:

[tex]\displaystyle\tt\longrightarrow L=\lim_{x\to 2}\dfrac{3^{x}\cdot 3^2-81}{(3^2)^x-(3^2)^2}[/tex]

[tex]\displaystyle\tt\longrightarrow L=\lim_{x\to 2}\dfrac{3^{x}\cdot 3^2-81}{(3^x)^2-(3^2)^2}[/tex]

[tex]\displaystyle\tt\longrightarrow L=\lim_{x\to 2}\dfrac{9(3^{x}-9)}{(3^x)^2-(3^2)^2}[/tex]

[tex]\displaystyle\tt\longrightarrow L=\lim_{x\to 2}\dfrac{9(3^{x}-9)}{(3^x+9)(3^x-9)}[/tex]

[tex]\displaystyle\tt\longrightarrow L=\lim_{x\to 2}\dfrac{9}{3^x+9}[/tex]

Substituting x = 2, we get:

[tex]\displaystyle\tt\longrightarrow L=\dfrac{9}{3^2+9}[/tex]

[tex]\displaystyle\tt\longrightarrow L=\dfrac{9}{9+9}[/tex]

[tex]\displaystyle\tt\longrightarrow L=\dfrac{9}{18}[/tex]

[tex]\displaystyle\tt\longrightarrow L=\dfrac{1}{2}[/tex]

Therefore:

[tex]\displaystyle\tt\longrightarrow \lim_{x\to 2}\dfrac{3^{x+2}-81}{9^x-9^2}=\dfrac{1}{2}[/tex]

Which is our required answer.

Answer:

[tex]\displaystyle\tt\hookrightarrow \lim_{x\to 2}\dfrac{3^{x+2}-81}{9^x-9^2}=\dfrac{1}{2}[/tex]

Learn More:

Some Standard limits.

[tex]\displaystyle\tt 1.\:\: \lim_{x\to0}\sin(x)=0[/tex]

[tex]\displaystyle\tt 2.\:\: \lim_{x\to0}\cos(x)=1[/tex]

[tex]\displaystyle\tt 3.\:\: \lim_{x\to0}\dfrac{\sin(x)}{x}=1[/tex]

[tex]\displaystyle\tt 4.\:\: \lim_{x\to0}\dfrac{\tan(x)}{x}=1[/tex]

[tex]\displaystyle\tt 5.\:\: \lim_{x\to0}\dfrac{1-\cos(x)}{x}=0[/tex]

[tex]\displaystyle\tt 6.\:\: \lim_{x\to0}\dfrac{\sin^{-1}(x)}{x}=1[/tex]

[tex]\displaystyle\tt 7.\:\: \lim_{x\to0}\dfrac{\tan^{-1}(x)}{x}=1[/tex]

[tex]\displaystyle\tt 8.\:\: \lim_{x\to0}\dfrac{\log(1+x)}{x}=1[/tex]

[tex]\displaystyle\tt 9.\:\: \lim_{x\to0}\dfrac{e^{x}-1}{x}=1[/tex]

[tex]\displaystyle\tt 10.\:\: \lim_{x\to0}\dfrac{(1+x)^n-1}{x}=n[/tex]

[tex]\displaystyle\tt 11.\:\: \lim_{x\to0}(1+x)^{\dfrac{1}{x}}=e[/tex]

[tex]\displaystyle\tt 12.\:\: \lim_{x\to\infty}\bigg[1+\dfrac{1}{x}\bigg]^{x}=e[/tex]

[tex]\displaystyle\tt 13.\:\: \lim_{x\to0}\dfrac{a^x-1}{x}=ln\:a[/tex]