Verified answer

Answer:

[tex]\boxed{\sf \: \: \displaystyle \lim_{x \to 0}\sf \: \dfrac{ e^{2x} - (1 + x)^7 }{ ln(1 + x) } \: = \: - \: 5 \: \: }\\ \\ [/tex]

Step-by-step explanation:

Given expression is

[tex]\sf \: \displaystyle \lim_{x \to 0}\sf \: \dfrac{ e^{2x} - (1 + x)^7 }{ ln(1 + x) } \\ \\ [/tex]

If we substitute directly x = 0, we get

[tex]\sf \: = \: \dfrac{ e^{0} - (1 + 0)^7 }{ ln(1 + 0) } \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{1 - 1}{ln(1)} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{0}{0} \\ \\ [/tex]

which is indeterminant form.

So, Consider again

[tex]\sf \: \displaystyle \lim_{x \to 0}\sf \: \dfrac{ e^{2x} - (1 + x)^7 }{ ln(1 + x) } \\ \\ [/tex]

can be rewritten as

[tex]\sf \: = \: \displaystyle \lim_{x \to 0}\sf \: \dfrac{ \dfrac{ e^{2x} - (1 + x)^7}{x} }{ \dfrac{ln(1 + x)}{x} } \\ \\ [/tex]

[tex]\sf \: = \: \displaystyle \lim_{x \to 0}\sf \: \dfrac{ e^{2x} - (1 + x)^7 }{x} \\ \\ [/tex]

[tex]\sf \: = \: \displaystyle \lim_{x \to 0}\sf \: \dfrac{( e^{2x} - 1) - [ (1 + x)^7 - 1]}{x} \\ \\ [/tex]

[tex]\sf \: = \: \displaystyle \lim_{x \to 0}\sf \: \dfrac{e^{2x} - 1}{x} - \displaystyle\lim_{x \to 0}\sf \dfrac{ {(1 + x)}^{7} - 1}{x} \\ \\ [/tex]

[tex]\sf \: = \: \displaystyle \lim_{x \to 0}\sf \: \dfrac{e^{2x} - 1}{2x} \times 2 - \displaystyle\lim_{1 + x \to 1}\sf \dfrac{ {(1 + x)}^{7} - 1}{1 + x - 1} \\ \\ [/tex]

We know,

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:\displaystyle\lim_{x \to 0}\sf \dfrac{ {e}^{x} - 1}{x} = 1 \qquad \: \\ \\& \qquad \:\sf \:\displaystyle\lim_{x \to a}\sf \dfrac{ {x}^{n} - {a}^{n} }{x - a} = {na}^{n - 1} \end{aligned}} \qquad \: \\ \\ [/tex]

So, using these results, we get

[tex]\sf \: = \: 1 \times 2 - 7 \times {(1)}^{7 - 1} \\ \\ [/tex]

[tex]\sf \: = \: 2 - 7 \\ \\ [/tex]

[tex]\sf \: = \: - 5 \\ \\ [/tex]

Hence,

[tex]\implies\sf \: \sf \: \displaystyle \lim_{x \to 0}\sf \: \dfrac{ e^{2x} - (1 + x)^7 }{ ln(1 + x) } \: = \: - \: 5\\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered} \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sin^{ - 1} x}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {tan}^{ - 1} x}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to \: a} \frac{ {x}^{n} - {a}^{n} }{x - a} = {na}^{n - 1} }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]

[tex]{\quad \huge{\underline{\bf{Limits\:\:and\:\:their\:\:Indeterminate\:\:forms}}}}[/tex]

Limit at a point actually is not the actual value of the function or the limand, or simply all the big big & sometimes even small mathematical expressions. If you have a simple limit, which is solvable without any manipulation, it's said the direct substitution. For example, we've the limit [tex]{\displaystyle \lim_{x\to 0}(x^{2}+1)}[/tex], if we just put x = 0, we get our limit as 1. But wait, didn't I told that it's not actually the value of the function? Yes, I did, so let's understand more deeper, first of all here x → 0, doesn't implies that x = 0, rather it means that x is approaching 0, or tending to 0, Now you've a human mind, so imagine the real number line in your mind, we -∞ at the left most side, +∞ at the right most side and 0 at the centre.So now we're approaching zero, we have two ways to be able to do so. Either we approach it from left to right or from right to left, so it here the limit being 1 means, if we approach zero from left to right, the function gets closer and closer to 1, same from right to left. Now we define two different limits, the left one, and the right one, more commonly known as the left hand limit(LHL) & the right hand limit(RHL). In left hand limit, we use the symbol, [tex]{x\to 0^{-}}[/tex] for LHL and [tex]{x\to 0^{+}}[/tex] for RHL. Now try the value on 0 itself, it's still 1, so evaluating a limit, just put the value approaching in it's variable occupied. So what about LHL & RHL? We change the variables to evaluate those limits? Ok, but how? Now as we know that x is not actually 0, and in case of LHL, it's approaching zero from left to right, implies it's less than zero, so we define, a variable [tex]{h}[/tex] such that, [tex]{x=0-h}[/tex], so the limand becomes, [tex]{h^{2}+1}[/tex], now as we've changed thr variable, the approaching value should also be changed, as for it, as we choosed x = -h, we put the approaching value of x, i.e 0 and we get the approaching value of h, which is 0 as well, now the LHL is [tex]{\displaystyle \lim_{h\to 0}(h^{2}+1)}[/tex] which gives 1, if you calculate RHL by the same approach and [tex]{x=0+h}[/tex], we get 1, so LHL = RHL = L(the limit, say), but what if LHL ≠ RHL or LHL = RHL ≠ L? In this case the limit doesn't exist, how? Let me explain, consider the function 1/(x-5), the LHL of the function as x → 5, is -∞, and RHL as x → 5 is +∞, now take the real number in imagination again? And the limit at 5 itself is not defined. Image all these three points approaching their respective limits, are they converging to the same point? No, so the limit doesn't exists (It was not the mathematical way, It was informal).

Indeterminate Forms:- These usually occur when the limand is a rational function say f(x)/g(x), and these mainly are 0/0, ∞/∞, ∞•∞, 0•∞, [tex]{1^{\infty}}[/tex] etc. If the limit exists, we can solve the limit by removing the factors or expressions which are the cause of it. There are a lot of methods, like Rationalization, factorisation, etc. But we still first use direct substitution, to check if the indeterminate form exists or not. A very method of evaluating indeterminate forms of rational type is L'Hopital's rule, we just keeps on differentiating both the numerator and denominator till the indeterminate form gets removed and then direct substitution and boom!

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Let's say the limit here given is L, then here L occupies 0/0 indeterminate form. So, using L'Hopital's rule;

[tex]{:\implies L=\displaystyle \lim_{x\to 0}\dfrac{2e^{2x}-7(1+x)^{6}}{\dfrac{1}{1+x}}}[/tex]

[tex]{:\implies L=1(2-7)}[/tex]

[tex]{\quad \qquad \therefore L=-5}[/tex]