Method 1 :
Once you take $x$ out of both Numerator and denominator of the expression $(x+6)/(x+1)$, We find that the Given limit is in $1^{∞}$ indeterminate form. Hence, We can directly apply the following formula (Easy to prove)
$\lim_{x \rightarrow a} f^g = e^{\lim_{x \rightarrow a} (f-1)g}$
Note that, To apply this Formula, $1/(f-1)$ must approach to $∞$ Which We have already seen Or can see that it does (It can be implied from the fact that the Given limit is in $1^{∞}$ indeterminate form)
Hence, We have our answer by applying this formula and that will be $e^5$ (Just a matter of putting values in the formula)
Method 2 :
Indeterminate forms of the form $1^∞,0^0,∞^0$ can be solved by Taking Logarithm(Taking Natural log makes calculation little less) both sides of the equation. (NOTE that $∞^∞, 0^∞, 0^{- ∞}, 1/∞,0/∞,∞/0$ are not considered indeterminate forms..)
Proof : $\lim_{x \rightarrow a}\,\, f^g = e^{\lim_{x \rightarrow a} (f-1)g}$
We know $\lim_{x \rightarrow a}\,\, (1+1/f)^f = e$ .....$Eq(1)$ Where $ \lim_{x \rightarrow a} \,\,f \rightarrow \,\, ∞ $ (We can prove it by taking Natural log both sides and then solve it...Try it)
Now let's get back to our equation to prove i.e. $\lim_{x \rightarrow a}\,\, f^g = e^{\lim_{x \rightarrow a} (f-1)g}$
Where $ \lim_{x \rightarrow a} \,\,f \rightarrow \,\, 1 $ and $ \lim_{x \rightarrow a} \,\,g \rightarrow \,\, ∞ $ (Since it is $1^∞$ form)
$\lim_{x \rightarrow a}\,\, f^g$ = $\lim_{x \rightarrow a}\,\, (1+f-1)^g $
= $\lim_{x \rightarrow a}\,\, (1+ \,\,\frac{1}{(\frac{1}{f-1})})^g $
= $\lim_{x \rightarrow a}\,\, (1+ \,\,\frac{1}{(\frac{1}{f-1})})^{g\frac{f-1}{f-1}} $
= $\lim_{x \rightarrow a}\,\, [(1+ \,\,\frac{1}{(\frac{1}{f-1})})^{1/(f-1)}]^{g(f-1)} $
= $\lim_{x \rightarrow a}\,\, [(1+ \,\,\frac{1}{(\frac{1}{f-1})})^{1/(f-1)}]^{\lim_{x \rightarrow a} g(f-1)} $
= $e^{\lim_{x \rightarrow a} (f-1)g}$
//Hence Proved
Useful formula to remember for $1^∞$ form. Remembering this will suffice for ALL $1^∞$ form.
