Continuity of $\tan\left(\frac{\pi}{4}+\ln x\right)^{\frac{1}{lnx}}$

Question

( First of all, apologie for my Math-Jax. I am still not very good at it. )

I am new to limits and continuity and was trying to solve a question:

If $f(x)= \tan\left(\frac{\pi}{4}+\ln x\right)^{\frac{1}{lnx}}$ is to be made continuous at $x=1$, then $f(1)$ should NOT be equal to: A) $e^2$ B) $e$ C)$\frac{1}{e}$ D) $e^{-2}$ ( ONE OR MORE OPTIONS MIGHT BE CORRECT )

Here is the photo of the question ( just to provide more clarity ):

I have 2 questions:

1) Why can't we simply put in $x=1$ in the function? The tangent term will become unity and 1 raised to the power whatever will always be 1. But from graph, $f(1)$ is not equal to $1$! Why is that?

2) From the graph itself, the function is continuous at $x=1$. Then why is the question saying " to be made continuous "?

It would be pretty cool, if you could do this without using the graph.

Answer 1

The function is continous at $1$ if $f(1) = \lim_\limits{x\to 1} f(x)$

$y = \lim_\limits{x\to 1} (\tan (\frac {\pi}{4}+\ln x)^{\frac {1}{\ln x}}\\ \ln y = \lim_\limits{x\to 1} \frac {\ln(\tan (\frac {\pi}{4}+\ln x))}{\ln x}$

Now you can apply l'Hopital's rule